The volume of the 1 km² body of water is 10,000,000 cubic meters (m³). The correct unit for volume is m³.
A measurement of three-dimensional space is volume. It is frequently expressed quantitatively using SI-derived units, as well as several imperial or US-standard units.
The area of the body of water is given as 1 km². To find the volume, we need to know the depth of the water. Let's assume the average depth of the body of water is 10 meters.
The volume of the body of water is:
Volume = Area x Depth
Volume = 1 km² x 1000 m/km x 10 m
Volume = 10,000,000 m³
Therefore, the volume of the 1 km² body of water is 10,000,000 cubic meters (m³). The correct unit for volume is m³.
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QX←→ is a tangent to circle R at point Q. Circle R with a chord and tangent. Point Q at 4 o clock on the circle. Point X at 2 o clock is outside the circle. Secant Line R Q and tangent line Q X. Angle XRQ is labeled 56 degrees. What is m∠RXQ ?
If Secant Line RQ and tangent line QX, angle XRQ is labeled 56 degrees, then the measure of angle RXQ is 34 degrees.
Since the line QX is tangent to the circle R, we know that angle QXR is a right angle. We also know that angle XRQ is labeled as 56 degrees. Therefore, we can find angle RXQ using the fact that the sum of the angles in a triangle is 180 degrees.
We have:
m∠RXQ + m∠QXR + m∠XRQ = 180 degrees
Since m∠QXR is a right angle (90 degrees), we can substitute:
m∠RXQ + 90 degrees + 56 degrees = 180 degrees
Simplifying:
m∠RXQ = 34 degrees
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the physician orders digoxin 0.25 mg po daily. the pharmacy supplies the following medication. the dosage strength of the digoxin can be expressed as: ? m g 1 t a b l e t
To calculate the dosage strength of digoxin, which is supplied by the pharmacy in mg per tablet, the physician orders digoxin 0.25 mg po daily.
In other words, The physician ordered 0.25 mg of digoxin to be administered orally every day. The medication provided by the pharmacy is to be taken in tablet form. To calculate the amount of digoxin in each tablet, you need to divide the ordered dose by the amount of tablets.
The equation is:Dose Ordered / Tablets = Dose per tablet
Substitute the known values:Dose Ordered = 0.25 mgTablets = 1 tablet0.25 mg / 1 tablet = 0.25 mg per tablet.
Therefore, the dosage strength of the digoxin supplied by the pharmacy is 0.25 mg per tablet.
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Please help super confused
The value of the function notations are;
a) d(20) = 10
b) d(0) = 0
c) d(20) < d(30)
d) d(0) = 0
e) d(30) = d(40)
f) d(t) = 5, when t = 5
How to Interpret Function Notation?
The notation f(x) defines a function named f. This is read as “y is a function of x.” The letter x represents the input value, or independent variable. The letter y is replaced by f(x) and represents the output value, or dependent variable.
a) d(20)
The value of distance (d) when the time is 20 is; 10
b) d(0)
The value of distance (d) when the time is 0 is; 0
c) d(20) = 10
d(30) = 20
Thus;
d(20) < d(30)
d) The value of the time at d = 0 is 0. Thus;
d(0) = 0
e) d(30) = 20
d(40) = 20
Thus; d(30) = d(40)
f) d(t) = 5
The value of t at d = 5 is;
t = 5
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Given that the measurement is in centimeters, find the area of the circle to the nearest tenth. (use 3.14 for π) circle with a radius of 3 cm
The area of the circle to the nearest tenth is 28.3 square centimeters.
To find the area of a circle with a given radius, we use the formula
A = π[tex]r^2,[/tex]
where A is the area and r is the radius.
In this case, the radius is 3 cm, so we can substitute it into the formula to get:
A = 3.14 x [tex]3^2[/tex]
Simplifying this equation, we get the following:
A = 3.14 x 9
A = 28.26
To round this to the nearest tenth, we look at the digit in the hundredth place, 6. Since 6 is greater than or equal to 5, we round up the number in the tenth place, which is 2. Therefore, the final answer is:
A ≈ 28.3 [tex]cm^2[/tex]
So, the area of the circle to the nearest tenth is 28.3 square centimeters.
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3/10 Students go for the music class and 2/3 students go for the dance class. which class has more students?
Answer: The dance class has more students
Step-by-step explanation:
3 divided by 10 = 0.3
2 divided by 3 = 0.66
Since 0.66 is more than 0.3, we can say that more students entered the dance class than the music class.
Please help superrr confused :(
Step-by-step explanation:
Week 1, you eat 10 out of 95
Week 1 r ( 1) = 95 - 10(1) =85
Week 5 r(5) = 95 - 10r
=95 - 10(5)
=95 - 50
=45lbs
Week 8 r(8) = 95 - 10.r
=95 - 10(8)
=95 - 80
= 15 lbs
Week w r(w) = 95 - 10.w
= 95 - 10W (lbs)
f(r) = 95 - 10r
Where r is the number of weeks.
r(w) =35 means that at the end of w weeks, there were 35 candies left.
If the original candy was 95 and 35 was left after w weeks
r(w) = 95 - 10w
35 = 95 - 10w
Subract 95 from both sides
-60 = - 10w
Divide both sides by - 10
6= w
The solution means that 35 candies were left after w weeks and w represents 6weeks.
(you can c compare my answer with a second response)
Find each quotient using synthetic division. (m^(4)-7m^(3)-39m^(2)-28m-3)-:(m+3)
The quotient using a synthetic method of division is m³ - 10m² - 9m - 1
How to evaluate the quotient using a synthetic methodThe quotient expression is given as
(m⁴ -7m³ -39m² - 28m - 3) divided by m + 3
Using a synthetic method of quotient, we have the following set up
-3 | 1 -7 -39 -28 -3
|__________
Bring down the first coefficient, which is 1:
-3 | 1 -7 -39 -28 -3
|__________
1
Multiply -3 by 1 to get -3, and write it below the next coefficient and repeat the process
-3 | 1 -7 -39 -28 -3
|____-3__ 30__27__3____
1 -10 -9 -1 0
So, the quotient is m³ - 10m² - 9m - 1
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a function is said to be differentiable at if exists. for some -values the derivative may not exist and we say that the function is not differentiable there. at which of the following locations is a function not differentiable? discontinuity cusp horizontal tangent line vertical tangent line
The location at which a function is not differentiable is a Vertical tangent line
A function may not be differentiable at some points. Such points are known as non-differentiable points. Let's take a look at each of the given terms to figure out the non-differentiable points. Discontinuity: A discontinuity is when a function's graph is interrupted by a break or hole.
It occurs when a function is undefined at a certain point. It may be classified into three categories: removable, jump, and infinite. Functions may not be differentiable at removable discontinuities but are differentiable at jump and infinite discontinuities. A discontinuous point is not the same as a non-differentiable point because it may be differentiable at other points of the function.
Cusp: A cusp is a sharp corner formed by a curve. It happens when the slope of the function approaches infinity. The curve is not differentiable at the cusp. Horizontal Tangent Line: When the slope of a function approaches zero, it creates a horizontal tangent line. The function may or may not be differentiable at this point depending on the shape of the graph.
It may be differentiable or not differentiable. Therefore, it is not a non-differentiable point. Vertical Tangent Line: When a function's slope approaches infinity, it creates a vertical tangent line. The function is non-differentiable at this point. A vertical tangent line is always a non-differentiable point.
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I need help this question is so confusing!
The bearing of B from C is calculated to be 115 degrees
The bearing of D from B is calculated to be 213.65 degrees
How to find the bearingsThe bearing from B to C is worked using SOH CAH TOA
Sin = opposite / hypotenuse - SOH
Cos = adjacent / hypotenuse - CAH
Tan = opposite / adjacent - TOA
The direction of movements describes a right triangle of
opposite = 30 km
adjacent = 30 km
The bearing is calculated using tan, TOA let the angle be x
tan x = Opposite / Adjacent
tan x = 30 / 30
tan x = 1
x = arc tan 1
x = 45 degrees
The bearing is 90 + 45 = 115 degrees
The bearing of D from B
x = arc tan (45/30)
x = 56.31
90 - 56.31 = 33.65
180 + 33.65 = 213.65 degrees
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Of the 90 families in our barangay,60 are engaged in farming and the rest are in fishing. What percent of the families are engaged in farming?
Answer:
66.67%
Step-by-step explanation:
There are two ways to solve this problem depending on which way you like. Percentages are based on a 0-100 system and thus you can start by dividing 100/90. This will give you an amount of 1 family. We are looking for 60 families, so multiply that value by 60 to get how much of a percentage 60 families is.
The other way is to divide 90 by 60, and then multiply that result by 100 to give you a percent.
Regardless of your preferred method, both answers are the same.
Which algebraic expression is equivalent to -2(4x - 5y - 5x)?
Answer: -2x-10y
Step-by-step explanation:i'm in the 8th grade so i know im right
In a box of 7 different banknotes 1, 2, 5, 10, 20, 50, 100 dollars. You pick at random 2 notes from this box. Let X is the larger number in these two notes, and Y be the smaller one.
(a) Find the distributions of X, Y and their joint distribution.
(b) Find the distribution of Z=X-Y.
(c) Compute E(X), Var(X), E(Z) and Var(Z).
The distribution of X is P(X = x) = 1/7 for x = 1, 2, 5, 10, 20, 50, 100 and Distribution of Y is P(Y = y) = 1/6 for y = 1, 2, 5, 10, 20, 50, 100, y ≠ X and their joint distribution is P(X = x, Y = x) = 0.
Find the joint distribution?(a) The possible values of X and Y are as follows:
(i) X can be any of the 7 banknotes.
(ii) Y can also be any of the 7 banknotes, except the one that was already picked for X.
Since each of the 7 banknotes is equally likely to be picked, the probability distributions of X and Y are both uniform distributions over the set {1, 2, 5, 10, 20, 50, 100}:
P(X = x) = 1/7 for x = 1, 2, 5, 10, 20, 50, 100
P(Y = y) = 1/6 for y = 1, 2, 5, 10, 20, 50, 100, y ≠ X
The joint distribution of X and Y can be computed as follows:
P(X = x, Y = y) = P(X = x) × P(Y = y|X = x)
Since the second banknote must be smaller than the first, we have:
P(Y = y|X = x) = 1/(6 - 1) = 1/5, for y ≠ x
And for y = x:
P(Y = x|X = x) = 0
Therefore:
P(X = x, Y = y) = 1/7 × 1/5 = 1/35, for x ≠ y
P(X = x, Y = x) = 1/7 × 0 = 0
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Given the equation 6x + 18 = 72:
Part A: Write a short word problem about a purchase made to illustrate the equation. (6 points)
Part B: Solve the equation showing all work. (4 points)
Part C: Explain what the value of the variable represents. (2 points)
If point A is the starting position, how high Is a rider after 3 seconds?
After 3 seconds, the rider is still at their starting height of 0 metres.
The height of a rider after 3 seconds can be calculated using the equation h(t) = v₀t - (1/2)gt², where h(t) is the height of the rider in metres at time t, v₀ is the initial velocity in metres per second, and g is the acceleration due to gravity in metres per second squared.
Assuming v₀ is 0, since the rider is just starting, and g is 9.81, then the height of the rider after 3 seconds is h(3) = 0 - (1/2)(9.81)(3²) = -44.355. Since this is a negative number, the rider is still at their starting height of 0 metres after 3 seconds.
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Suppose given a representation of the symmetric group S3 on a vector space V. Let x and y denote the usual generators for S3. (a) Let u be a nonzero vector in V. Let v = u + xu + xều and w = u + yu. By analyzing the G-orbits of v, w, show that V contains a nonzero invariant subspace of dimension at most 2. (b) Prove that all irreducible two-dimensional representations of G are isomorphic, and determine all irreducible representations of G
The orbits of v and w imply a nonzero invariant subspace of dimension at most 2. All 2-dimensional irreducible representations of S3 are isomorphic and can be determined by the character table.
To show that V contains a nonzero invariant subspace of dimension at most 2, we analyze the G-orbits of v and w. Since x and y are the generators for S3, we can see that xv = x(u + xu + xều) = x²u + xều = u and yw = y(u + yu) = u + y²u = u. This implies that the G-orbits of v and w are the same, so V contains a nonzero invariant subspace of dimension at most 2. To prove that all irreducible two-dimensional representations of G are isomorphic, we must show that they all have the same character table. To do this, we can use the fact that the character table is an invariant of a representation, meaning that it is the same for all isomorphic representations. Therefore, all irreducible two-dimensional representations of G have the same character table, which can be determined by examining the character table for S3.
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Jane has a pre-paid cell phone with Splint. She can't remember the exact costs, but her plan has a
monthly fee and a charge for each minute of calling time. In June she used 350 minutes and the cost
was $177.00. In July she used 900 minutes and the cost was $397.00.
A) Express the monthly cost C as a function of x, the number of minutes of calling time she used.
Answer: c(x) = 4x +
syntax error.
B) If Jane used 622 minutes of calling time in August, how much was her bill?
Answer: $
On Sunday, the owners of the Middletown Café are giving away muffins. The owners budgeted $75 to spend on muffins for the event, and each muffin costs $0.82. The inequality 75≥0.82m 75 ≥ 0 . 82 , where m is the number of muffins, represents the situation. How many customers could possibly get a muffin? Select all that apply.
Answer:
We can solve the inequality for m to find the maximum number of muffins that can be purchased within the budget of $75:
75 ≥ 0.82m
Divide both sides by 0.82:
m ≤ 91.46
Since m must be a whole number, the maximum number of muffins that can be purchased is 91. Therefore, 91 customers could possibly get a muffin.
Step-by-step explanation:
please help if you can I will give brainliest- Black tape is used to create the lines and circles for a basketball court. How much tape is used in all? Use π=3.14.
Answer:
see below.
Step-by-step explanation:
So first, find the perimeter.
94*2= 188
50*2=100
100+188=288ft <----This is our perimeter
Next, find the length of the center line.
It is 50 feet.
Next, we are going to figure out the circumference of the big half circle.
C=[tex]\pi[/tex]d
C=(3.14)(44)=138.16 <------circumference of a WHOLE circle.
Divide 138.16 by 2 to get the half circle
138.16/2=69.08
There are 2 half circles, so multiply 69.08 by 2 to get 138.16
Next, we need to find the circumference of the center circle.
C=(3.14)(12)=37.68
Finally, add all the totals together
288+50+138.16+37.68=513.84 FEET <---dont forget units.
The following system of equations are given:
3x+z+y=8
5y-x=-7
3z+2x-2y=15
4x+5y-2z=-3
a. Is it possible to solve for any of the variables using only Equation #1 and Equation #2? Explain your answer. If possible, solve for the variables using only equations #1 and #2.
b. Is it possible to solve for any of the variables using only Equation #1, Equation #2, and Equation #3? Explain your answer. If possible, solve for the variables using only equations #1, #2, and #3.
c. If you found solutions in part b, do these solutions also hold for Equation #4?
The solution for the system of equations using Equations #1, #2, and #3 is x=-55, y=-9.6, z=39.2.
What is Linear Equation ?
Linear equation can be defined as equation in which highest degree is one.
We can solve for one variable using Equations #1 and #2, but not for all three variables. From Equation #2, we can solve for x in terms of y as x=5y+7. Substituting this expression for x into Equation #1 gives:
3(5y+7)+z+y=8
Simplifying this equation, we get:
16y+z=-13
We still have two unknowns, y and z, so we cannot solve for any variable using only Equations #1 and #2.
b. It is possible to solve for all three variables using Equations #1, #2, and #3. We can use the following steps to solve for the variables:
Use Equation #2 to solve for x in terms of y as x=5y-7.
Substitute this expression for x into Equation #3 to get:
3z+2(5y-7)-2y=15
Simplifying this equation, we get:
13y+3z=29
Substitute the expression for x from Step 1 into Equation #1 to get:
3(5y-7)+z+y=8
Simplifying this equation, we get:
16y+z=29
Solve for z in terms of y by subtracting the equation from Step 3 from the equation from Step 2:
(13y+3z)-(16y+z)=29-(-13)
Simplifying this equation, we get:
-3y+2z=42
Solve for z in terms of y by adding twice the equation from Step 1 to the equation from Step 4:
2(5y-7)+(-3y+2z)=42
Simplifying this equation, we get:
7y+2z=56
Solve for y by adding the equation from Step 4 to twice the equation from Step 5:
(7y+2z)+2(-3y+2z)=56+84
Simplifying this equation, we get:
5z=196
So, z=39.2. Substituting this value for z into the equation from Step 4 gives:
-3y+2(39.2)=42
Solving for y, we get y=-9.6. Finally, substituting the values for y and z into the equation from Step 1 gives:
x=5(-9.6)-7=-55.
Therefore, the solution for the system of equations using Equations #1, #2, and #3 is x=-55, y=-9.6, z=39.2.
c. To check if the solution found in part b holds for Equation #4, we substitute the values of x, y, and z into Equation #4 and see if the equation is satisfied:
4(-55)+5(-9.6)-2(39.2)=-3
Simplifying this equation, we get:
-220-48-78.4=-3
This equation is not satisfied, so the solution found in part b does not hold for Equation #4. Therefore, the system of equations does not have a unique solution.
Therefore, the solution for the system of equations using Equations #1, #2, and #3 is x=-55, y=-9.6, z=39.2.
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Write a decimal and a fraction for the shaded part of the diagram.
A decimal and fraction and shaded part of the first diagram second diagram is 100/100 or 1 and 0.63 or 63/100 respectively.
What is fraction?
A fraction is a way of expressing a part of a whole, or a part of a group, by using two numbers separated by a line. The number above the line is called the numerator, and the number below the line is called the denominator.
The first 10 by 10 matrix is fully colored, which means all 100 cells are colored. Therefore, the fraction of colored cells is:
100/100 = 1
The decimal representation of this fraction is simply 1.
For the second 10 by 10 matrix, the first 6 columns are fully colored, which means there are 6 x 10 = 60 colored cells. The 7th column has only the first 3 rows colored, which means there are 3 colored cells in this column. Therefore, the total number of colored cells is:
60 + 3 = 63
The fraction of colored cells is:
63/100
To convert this fraction to a decimal, we can divide the numerator by the denominator:
63 ÷ 100 = 0.63
Therefore, the shaded part of the second matrix is 0.63 or 63/100.
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PLEASE SHOW WORK!!!!!!!!!
The solid shown here is a cube. Count the number of faces, edges, and vertices. Remember, you can use the formula V – E + F = 2 to make sure that you counted correctly.
Vertices
Edges
Faces
In the given cube, the required data is as follows:
Faces = 6
Edges = 12
Vertices = 8
What is a cube?A cube is a solid three-dimensional form with six square faces that all have the same length sides. It is one of the five platonic solids and is also referred to as a regular hexahedron.
Six square faces, eight vertices, and twelve edges make up the form.
Here in the question as asked,
Faces = 6
Edges = 12
Vertices = 8
Now to prove that we are correct,
V -E + F =2
= 8 - 12 + 6
=2
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Can someone write this 0.698, 0.2, 0.099, 0.18 in order please?
Answer:
0.099, 0.18, 0.2, 0.698.
Step-by-step explanation:
find the lateral surface area of a square 14 mm 13mm 6 mm
The lateral surface area of the square is 784mm².
What is square?Having four equal sides, a square is a quadrilateral. There are numerous square-shaped objects in our immediate environment. Each square form may be recognised by its equal sides and 90° inner angles. A square is a closed form with four equal sides and interior angles that are both 90 degrees. Numerous different qualities can be found in a square.
If the square has a side length of 14 mm, then all sides are equal in length.
The lateral surface area of a square is simply the perimeter multiplied by the height of the shape. Since a square has equal length and width, the perimeter of the square is simply 4 times its side length. Therefore, we have:
Perimeter = 4 × 14 mm = 56 mm
The height of the shape is the same as the length of any of its sides, which is also 14 mm. Therefore, we have:
Lateral Surface Area = Perimeter × Height = 56 mm × 14 mm = 784 mm^2
Therefore, the lateral surface area of the square is 784mm².
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find the lateral surface area of a square box with dimensions 14 mm, 13mm and 6 mm.
the domain for the first input variable to predicate t is a set of students at a university. the domain for the second input variable to predicate t is the set of math classes offered at that university. the predicate t(x, y) indicates that student x has taken class y. sam is a student at the university and math 101 is one of the courses offered at the university. give a logical expression for each sentence. (a) sam has taken math 101. (b) every student has taken at least one math class. (c) every student has taken at least one class other than math 101. (d) there is a student who has taken every math class other than math 101. (e) everyone other than sam has taken at least two different math classes. (f) sam has taken exactly two math classes.
The logical expressions for each sentence can be written as follows:
(a) t(Sam, Math 101)
This expression states that Sam has taken Math 101.
(b) ∀x∃y t(x, y)
This expression states that for every student x, there exists a math class y such that the student x has taken the math class y.
(c) ∀x∃y (t(x, y) ∧ y ≠ Math 101)
This expression states that for every student x, there exists a class y such that the student x has taken the class y and the class y is not Math 101.
(d) ∃x∀y (t(x, y) ∧ y ≠ Math 101)
This expression states that there exists a student x such that for every math class y, the student x has taken the math class y and the math class y is not Math 101.
(e) ∀x∃y∃z (t(x, y) ∧ t(x, z) ∧ x ≠ Sam ∧ y ≠ z)
This expression states that for every student x, there exists two different math classes y and z such that the student x has taken the math classes y and z and the student x is not Sam.
(f) ∃y∃z (t(Sam, y) ∧ t(Sam, z) ∧ y ≠ z ∧ ∀w (t(Sam, w) → (w = y ∨ w = z)))
This expression states that there exists two different math classes y and z such that Sam has taken the math classes y and z and for every math class w, if Sam has taken the math class w, then the math class w is either y or z.
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Question 7 (3 points)
solve for c: -2=c/16
Question 7 options:
c = - 1/8
c = -32
c = 32
c = 8
The solution of the equation -2 = c/16 by making c the subject of equation is c = -32
How to solve an equation?An equation is an expression containing numbers and variables linked together by mathematical operations such as addition, subtraction, division, multiplication and exponents.
Given the equation:
-2 = c/16
We are to solve for c by making c the subject of equation. To solve for c, we multiply both sides of the equation by 16 and simplify. Therefore:
-2 * 16 = c/16 * 16
Simplifying:
c = -32
The solution of the equation -2 = c/16 is c = -32
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5, 6, 8, ___ , 15, ___
18, 20, 24, ___ , 38, ___
25, 28, 34, ___ , ___ , 70
55, 54, 51, 46, ___ , ___ , 19
82, 81, 78, ___ , 66
0 + 6 = ___+ 0
___ + 9 = 9 + 14
20 x ( 4 + ___) = ( 20 x 4 ) + (20 x 3)
9 + ( 6 + 5 ) = (9 + 6) + ___
10 x (___ + 6) = (10 x 8) + (10 x ___)
Answer:
5, 6, 8, 11, 15, 20
5+1=6+2=8+3=11+4=15+5=20
18, 20, 24, 30, 38, 48
18+2=20+4=24+6=30+8=38+10=48
55, 54, 51, 46, 39, 30, 19
55-1=54-3=51-5=46-7=39-9=30-11=19
82, 81, 78, 73, 66
82-1=81-3=78-5=73-7=66
0+6=6+0
6=6
14+9=9+14
23=23
20x(4+3)=(20x4)+(20x3)
20x7=80+60
140=140
9+(6+5)=(9+6)+5
9+11=15+5
20=20
10x(4+6)=(10x8)+(10x2)
100=100
Step-by-step explanation:
What is the value of sinD?
The value of sin(D) is 7/25 after the application of the Pythagoras theorem.
What is a Pythagoras theorem?The Pythagorean theorem is a fundamental theorem in geometry that describes the relationship between the sides of a right triangle. It claims that the hypotenuse's square length, which is the side that faces the right angle, is equivalent to the total of the squares of the lengths of the other two sides in a right triangle. The theorem can be formulated mathematically as:
c² = a² + b²
where, even the lengths for the remaining two sides (the legs) of the right triangle are a and b, and c is the length of the hypotenuse.
The Pythagorean theorem may be employed to determine the triangle's third side's length:
DE²= FD² + EF²
25² = 24² + EF²
625 = 576 + EF²
EF² = 49
EF = 7
Now, we can use the definition of sine to find sin(D):
sin(D) = opposite/hypotenuse = EF/DE = 7/25
Therefore, the value of sin(D) is 7/25.
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Question 7 What equation is parallel to y=-(1)/(4)x+5 and passes through (2,-3)?
The equation of the line that is parallel to y = -(1/4)x + 5 and passes through (2, -3) is y = -(1/4)x - 5/2.
To find the equation of the line that is parallel to the line y = −(1/4)x + 5 and passes through the point (2, −3), follow these steps:Step 1: Determine the slope of the given line.The slope-intercept form of the equation of the line is y = mx + b where m is the slope of the line. y = −(1/4)x + 5 is already in slope-intercept form, so its slope is −1/4. Step 2: Determine the slope of the line that is parallel to the given line. The slope of a line parallel to another line is the same as the slope of the given line. Therefore, the slope of the line we need to find is also −1/4. Step 3: Determine the y-intercept of the line we need to find. We already know that the line passes through the point (2, −3). To determine the y-intercept of the line, substitute x = 2 and y = −3 into the slope-intercept form of the equation of the line. −3 = −(1/4)(2) + b b = −3 + 1/2 = −5/2 Therefore, the y-intercept of the line we need to find is −5/2. Step 4: Write the equation of the line in slope-intercept form. The equation of the line we need to find is y = mx + b where m = −1/4 and b = −5/2. y = −(1/4)x − 5/2 is the equation of the line that is parallel to y = −(1/4)x + 5 and passes through (2, −3).Answer: The equation of the line that is parallel to y = -(1/4)x + 5 and passes through (2, -3) is y = -(1/4)x - 5/2.
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the shape is formed from two straight lines and two arcs. work out the total shaded area correct to the nearest 0.1cm^2.
Check the picture below.
so we're really looking for the area of a sector of a circle with 63° and a radius of 3, twice.
[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360} ~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ \theta =63\\ r=3 \end{cases}\implies A=\cfrac{(63)\pi (3)^2}{360} \\\\\\ A=\cfrac{63\pi }{40}\implies \stackrel{\textit{now let's double that}}{2\cdot \cfrac{63\pi }{40}}\implies \cfrac{63\pi }{20}\implies 9.9~cm^2[/tex]