Use the technique of linear regression to find the line of best fit for the given points. Round any intermediate calculations to no less than six decimal places, and round the coefficients to two decimal places.

(1,2)
, (2,5)
, (3,7)
, (4,2)
, (5,10)
, (6,10)
, (7,9)

Answer:

Propaganda: ideas or statements spread to promote one side's views and present an opposing side negatively.

Front: the front lines where combat between opposing armies occurs.

Free enterprise system: an economic system characterized by private or corporate ownership of capital goods.

Allied Forces: the military alliance of the USA, Britain, France, and other countries during WWII.

Axis Powers: Germany, Italy, Japan.

Using drawing tools parabola graph has been made and uploaded in answer for the given equation.

Standard form of parabola having vertical axis of symmetry given by:

[tex](y - k)^2 = 4p(x - h)[/tex]

where (h, k): vertex and p: distance between the vertex and the focus or directrix.

Comparing the given equation[tex]x = -1/8(y - 3)^2 + 1[/tex] with the standard form, we can see that the vertex is at (1, 3) and p = -1/32.

Since the parabola opens to the left, the focus is to the left of the vertex at a distance of p = -1/32 units. Thus, the focus is located at (-1/32, 3).

The directrix is a vertical line to the right of the vertex and is located at a distance of p = -1/32 units. Thus, the directrix is the vertical line x = 33/32.

To graph the focus and directrix, we can plot the vertex (1, 3) on the coordinate plane, draw a horizontal line through the vertex, and then plot the focus (-1/32, 3) to the left of the vertex and the directrix x = 33/32 to the right of the vertex.

Note that the parabola [tex]x = -1/8(y - 3)^2 + 1[/tex] is symmetric with respect to the vertical line passing through the vertex.

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Answer:

C.

Step-by-step:

The measure of angle A is congruent to the measure of angle P.

Answer:  a = [tex]\frac{2}{7}[/tex].

Step-by-step explanation:

    10a - 2 = 4a - 1a

    10a - 2 = 3a

10a -2 + 2 = 3a + 2

          10a = 3a + 2

   10a - 3a = 3a -3a + 2

            7a = 2

              a = [tex]\frac{2}{7}[/tex]

The vertices will be D = (-3,2) E=(1,2) F=(-2,0).

resizing an οbject is accοmplished thrοugh a change called dilatiοn. The οbjects can be enlarged οr shrunk via dilatiοn. A shape identical tο the sοurce image is created by this transfοrmatiοn. The size οf the fοrm dοes, hοwever, differ. A dilatatiοn οught tο either extend οr cοntract the οriginal fοrm. The scale factοr is a phrase used tο describe this transitiοn.

The scale factοr is defined as the difference in size between the new and οld images. An established lοcatiοn in the plane is the center οf dilatatiοn. The dilatiοn transfοrmatiοn is determined by the scale factοr and the centre οf dilatiοn.

Here the scale factοr is 1/2

Sο the vertices will be D = (-3,2)

E=(1,2)

F=(-2,0)

Hence the vertices will be D = (-3,2) E=(1,2) F=(-2,0).

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The false comparison is: 35 degrees centigrade cooler than 100 degrees Fahrenheit.

In mathematics and computer programming, an expression is a combination of one or more values, variables, operators, and functions that are evaluated to produce a result.

The comparison "35 degrees centigrade cooler than 100 degrees Fahrenheit" is false because 35 degrees centigrade is equivalent to 95 degrees Fahrenheit, which is actually warmer than 100 degrees Fahrenheit. Therefore, it is incorrect to say that 35 degrees centigrade is cooler than 100 degrees Fahrenheit.

Therefore, the false comparison is: 35 degrees centigrade cooler than 100 degrees Fahrenheit.

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The rock will be 125  feet above the beach at: The values of t are 3.911 or 1.28 and (b) The minimum height is h= 346 feet

Velocity is the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time

The given parameters are

h = -16t² + 79t + 50

In the equation put h = 125 to find t

125 = -16t² +79t+50

Rearrange the equation to have

16t²-79t -50 +125 = 0

this is also written as 16t² -79t + 75 =0

Solving for t we have

[-b±√b²-4ac]/2z

[79 ±√79²-4*16*75]/2*16

[(79 ±√6241-4800)] / 32

Simplify further to get

[79 ±√1441] / 32

(79 ±38) / 32

117/32 or 41/32

The values of t are 3.911 or 1.28

The minimum height of the rock at t = 1.28 is

h = -16(3.28)² + 79( 3.28)+ 50

h = -16(10.7584) + 79(3.28) +50

h = -  172.1344 +259.12 + 50

h= 346 feet

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The ball strikes the grοund after apprοximately 1.07 secοnds.

Time: a measured οr measurable periοd during which an actiοn, prοcess, οr cοnditiοn exists οr cοntinues.

Speed: the rate at which sοmething mοves, οperates, οr happens.

Distance: the amοunt οf space between twο pοints οr οbjects.

We can sοlve this prοblem using the kinematic equatiοns οf mοtiοn. We knοw the initial velοcity (u = 7.30 m/s), the initial height (h = 2.91 m), and we want tο find the time taken fοr the ball tο hit the grοund (t).

Let's use the kinematic equatiοn that relates the final pοsitiοn (s), initial pοsitiοn (h), initial velοcity (u), acceleratiοn (a), and time (t):

[tex]s = h + ut + (1/2)at^2[/tex]

Since the ball is thrοwn vertically dοwnwards, the acceleratiοn is [tex]-9.81 m/s^2[/tex] (negative because it is in the οppοsite directiοn tο the initial velοcity).

At the mοment the ball hits the grοund, its final pοsitiοn s is zerο. Therefοre, we can rewrite the equatiοn as:

[tex]0 = h + ut + (1/2)at^2[/tex]

Solving for t, we get:

[tex]t = [-u\±\sqrt{(u^2 - 2ah)} ] / a[/tex]

We take the positive solution, since time cannot be negative. Substituting the values we get:

[tex]t = [ -7.30 \± \sqrt{((7.30)^2 - 2(-9.81)(2.91))} ] / (-9.81)[/tex]

[tex]t = [ -7.30 \± \sqrt{ (53.25)} ] / (-9.81)[/tex]

t ≈ 1.07 s οr t ≈ 0.31 s

The negative sοlutiοn dοesn't make physical sense in this cοntext, sο we discard it. Therefοre, the ball strikes the grοund after apprοximately 1.07 secοnds.

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The features of the given function are;

y-intercept at (0,10)

vertical asymptote of x = -4

A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.” The input values make up the domain, and the output values make up the range.

The function is given as;

g(x) = f(x + 4) + 8

The domain of a function is a set of all possible input values while the range of a function is defined as a set of all possible output values.

A vertical asymptote of a graph is a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a.

Thus, vertical asymptote is x = -4.

The he feat will be at (0, 10)

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Answer:

B) 2≤x≤8.

Step-by-step explanation:

Add 6 to all sides of the inequality.

0+6≤3x-6+6≤18+6

6≤3x≤24

Divide all sides by 3.

2≤x≤8

The correct answer is B) 2≤x≤8.

Answer:

(36 × $7) - $15 = $252 - $15 = $237

The measure οf the third angle c is 20 degrees.

The triangle angle sum prοperty states that the sum οf the three interiοr angles οf a triangle is always equal tο 180 degrees. This prοperty hοlds true fοr all types οf triangles, whether they are acute, οbtuse, οr right-angled. The prοperty can be expressed as fοllοws:

Angle A + Angle B + Angle C = 180 degrees, where A, B, and C are the measures οf the interiοr angles οf the triangle.

38 + 122 + c = 180

Tο sοlve fοr c, we can simplify the left side οf the equatiοn:

160 + c = 180

Subtracting 160 frοm bοth sides, we get:

c = 20

Therefοre, the measure οf the third angle c is 20 degrees.

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Complete question:

Two angles in a triangle are 38° and 122°. Write the angle sum property to determine the measure of the third angle ∠c.

[Hint: Angle sum property = ∠a+ ∠b+ ∠c = 180°]

Answer:

50 sweets

Step-by-step explanation:

Let's work through the problem step by step:

The student starts with x sweets.

She gives away 20 sweets to her friends, so she is left with (x - 20) sweets.

One third of the remainder is equal to one fifth of the original number of sweets. In other words, (1/3)(x - 20) = (1/5)x.

To solve for x, we can start by multiplying both sides of the equation by 15 (the least common multiple of 3 and 5) to eliminate the fractions:

5(x - 20) = 3x

5x - 100 = 3x

2x = 100

x = 50

Therefore, the original number of sweets was 50.

Answer:

420 K x (60 cm³ / 70 cm³) x (7 / 18) = 294.29 K

Step-by-step explanation:

The perimeter of the triangle will be 36.

The value of x in the triangle will be 101°.

The value of y will be 3x

It should be noted that the total sum of the angles that are in a triangle is 180°. It should be noted that the triangle is a Equilateral Triangle. In this case, since one of the sides is 12, the perimeter will be:

= 12 + 12 + 12

= 36

The value of x in the triangle will be:

= 180° - (22 + 57)

= 180° - 79°

= 101°.

The value of y will be 3x as vertically opposite angles are equal.

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Exponential property can be used in various field or place like growing or decline of population, bacteria growth/decay , compound interest any many more.

Exponential property states that when we multiply  two power with same base we add the exponents.

                 [tex]a^{m} *b^{n } = (a*b)^{m + n}[/tex]

When we divide  two power with same base we subtract the exponents.

                  [tex]a^{m} \div b^{n} = (a \div b)^{m-n}[/tex]

When have to find the power of power we multiply the exponents.

                  [tex](a^{m} )^{n} = a^{m*n}[/tex]

Three most important use of exponential function is to calculate carbon dating of any soil, fossil foil etc, to calculate population growth , interest earn on an investment etc.

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Answer:

YOU HAVE TO USE THE VECTORS AS WELL MATRIX TO GET THE SOLUTION OF THIS QUESTION

Since the distance was measured to the nearest metre, the upper bound for the distance is 101 metres.

1. To calculate the lower bound for Kelly's average speed, we need to divide the lower bound distance by the upper bound time. Since the distance was measured to the nearest metre, the lower bound for the distance is   [tex]99[/tex] Metres.

Since the time was measured to the nearest hundredth of a second, the upper bound for the time is  [tex]10.53[/tex] seconds. Therefore, the lower bound for Kelly's average speed is:

[tex]99/10.53 = 9.411[/tex]  metres per second (to three decimal places)

2. To calculate the upper bound for the perimeter of the regular hexagon, we need to multiply the upper bound length of a side by 6.

Since the length was measured to the nearest millimetre, the upper bound for the length is [tex]3.601 cm[/tex]  (since 3.6005 cm would round up to 3.601 cm). Therefore, the upper bound for the perimeter is:

[tex]6 x\times3.601 = 21.606 cm[/tex]   (to three decimal places)

3. To calculate the upper bound for the area of the rectangle, we need to multiply the upper bounds for the length and width. Since the length is correct to the nearest cm, the upper bound for the length is 35.5 cm (since 34.5 cm would round up to 35 cm).

Since the width is correct to the nearest cm, the upper bound for the width is 26.5 cm (since 25.5 cm would round up to 26 cm). Therefore, the upper bound for the area is:

 [tex]35.5 \times 26.5 = 942.25 cm^2[/tex] (to two decimal places)

4. To calculate the lower bound for Kelly's average speed, we need to divide the upper bound distance by the lower bound time.

Since the time was measured to the nearest hundredth of a second, the lower bound for the time is 10.51 seconds. Therefore, the lower bound

(d) To calculate the lower bound for Kelly's average speed, we need to divide the lower bound of distance by the upper bound of time.

The lower bound of distance is 99.5m (since the measurement was rounded down to the nearest metre).

The upper bound of time is 10.525s (since the measurement was rounded up to the nearest hundredth of a second).

So, the lower bound for Kelly's average speed is:

speed = distance / time [tex]= 99.5 / 10.525 = 9.45260637[/tex] ...

We need to round this to two decimal places to match the precision of the time measurement, giving us:

speed = [tex]9.45 m/s[/tex]

Therefore, the figures on the calculator display are: 99.5 ÷ 10.525 = 9.45260637... ≈ 9.45.

The length of the rectangle is measured as 645 mm correct to the nearest 5 mm. This means that the actual length could be anywhere between 642.5 mm and 647.5 mm (since rounding up or down depends on the decimal value being greater or less than 0.5 respectively).

Similarly, the width of the rectangle is measured as 400 mm correct to the nearest 5 mm. This means that the actual width could be anywhere between 397.5 mm and 402.5 mm.

5. To calculate the lower bound for the area of the rectangle, we need to find the product of the smallest possible length and width.

Smallest possible length [tex]= 642.5 mm[/tex]

Smallest possible width  [tex]= 397.5 mm[/tex]

Area = length x width

Lower bound for area = [tex]642.5 mm x 397.5 mm = 255542.5 mm²[/tex]

Rounding this off to 3 significant figures, we get the final answer as 2.56 x 10^5 mm².

Therefore, the lower bound for the area of the rectangle is [tex]2.56 x 10^5[/tex]  mm².

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Answer:

1. c

2. b

3. B

Step-by-step explanation:

1. the hypotenuse is the longest side of the triangle.

2. think of it as the leg that helps form angle A (or forms right angle) that is NOT the hypotenuse

3. if "c" is the hypotenuse,, & b is the adjacent leg,, the opposite angle is across from the given angle which in this case is A

Answer:

the area where the grass will be planted is 25,920 square meters.

Step-by-step explanation:

To calculate the area where the grass will be planted, we need to multiply the length and breadth of the sports field.

Area = length x breadth

Area = 162m x 160m

Area = 25,920 square meters

Therefore, the area where the grass will be planted is 25,920 square meters.

Answer:

no

Step-by-step explanation:

The formula d=b2−4ac when solved for b = ± [tex]\sqrt{d +4ac}[/tex]. Option B

Subject of formula is defined as a variable that is being solved for in an equation.

It is the variable that is made to stand out in an equation. It is the variable that is made to stand alone on one end of the equality sign.

From the information given, we have that;

d=b2−4ac

Now, collect the like terms

d + 4ac = b²

Find the square root of both sides, we have that;

b = [tex]\sqrt{d + 4ac}[/tex]

Insert the plus-minus sign to show the interval, we have;

b = ± [tex]\sqrt{d +4ac}[/tex]

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The name of this circle is Circle R.

The name of the radius is RB.

The name of the diameter is RI or RD.

In Mathematics, a circle can be defined as a closed, two-dimensional curved geometric shape with no edges or corners. Additionally, a circle refers to the set of all points in a plane that are located at a fixed distance (radius) from a fixed point (central axis).

In Mathematics, a line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points. Additionally, a line segment typically has a fixed length.

In this context, we have the following names;

Circle R.

Radius = RB.

Diameter = RD or RI.

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Equations that have infinitely many solutions are:

B. 3x+8-x=3+5+2x

D. 5-4x+7= −2 (2x - 6)

An equation has infinitely many solutions when you try to solve the equation and you get a variable or a number equal to itself.

A. 3 (x + 4) = 2x - 7

3x + 12 = 2x - 7

3x -2x = -7 -12

x = -19

B. 3x+8-x=3+5+2x

3x -x - 2x =3 + 5 - 8

0 = 0

This equation has infinitely many solutions.

C. (4x − 8) = 2x − 4 + 12

4x − 2x =  − 4 + 12 + 8

2x = 16

x = 8

D. 5-4x+7= −2 (2x - 6)

5 - 4x + 7 = −4x + 12

-4x + 4x = 12 - 7 - 5

0 = 0

This equation has infinitely many solutions.

E. 4-3x - 8 = 2x + 7 - 5x

-3x - 2x + 5x = 7 + 8 - 4

0 = 11

Therefore, options B and D has infinitely many solutions.

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Answer: An array is a collection of items of same data type stored at contiguous memory locations.

Step-by-step explanation: There you go dude happy to help

Answer:Not dividing the entire side of the equation.

Not distributing properly using F.O.I.L.

Confusing the rules for adding and subtracting negative numbers with the rules for multiplying and dividing negative numbers.

Multiplying exponents with the base.

Step-by-step explanation:

Answer:

Coefficient, the coefficient is the number before the variable

Step-by-step explanation:

The mathematical word that describes both 2.25 and 6.75 in the expression 2.25x + 6.75x - 5.5 is "coefficients."

An expression contains one or more terms with addition, subtraction, multiplication, and division.

We always combine the like terms in an expression when we simplify.

We also keep all the like terms on one side of the expression if we are dealing with two sides of an expression.

Example:

1 + 3x + 4y = 7 is an expression.

3 + 4 is an expression.

2 x 4 + 6 x 7 – 9 is an expression.

33 + 77 – 88 is an expression.

We have,

In this expression,

2.25 and 6.75 are coefficients of the variables x, which means that they multiply the variable x.

The term 5.5 is a constant because it does not contain a variable.

Thus,

The mathematical word that describes both 2.25 and 6.75 in the expression 2.25x + 6.75x - 5.5 is "coefficients."

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The approximate value of the given logarithmic expression as required to be determined in the task content is; 1.528.

It follows from the task content; it is required that the value of the logarithmic expression is to be determined.

Therefore, we have;

let the result of the expression be x; so that we have;

log₈ (24) = x

x log (8) = log (24)

x = log (24) - log (8)

x = 1.528

Ultimately, the approximate value of the logarithmic expression in discuss is; 1.528.

Complete question: The correct expression is; log₈ (24) .

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Answer:

1) S=31.68 yd  A=174.23 [tex]yd^{2}[/tex]

2) S=36.65 in A=256.56 [tex]in^{2}[/tex]

3) S=8.9 ft A=13.35 [tex]ft^{2}[/tex]

4)S=17.28 in. A=51.84 [tex]in^{2}[/tex]

5)S=25.31 yd. A=126.54 [tex]yd^{2}[/tex]

6)S=3.4ft A=5.11 [tex]ft^{2}[/tex]

Step-by-step explanation:

Notes:

360 was divided by 1/2, making the denominator 180. This is why there is a 1/2 at the front.

You also do not need to simplify if you're using a calculator.

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

- r is the radius

- θ is [tex]\frac{angle\pi }{180}[/tex]

S= r θ

-r is the radius

- θ is [tex]\frac{angle\pi }{180}[/tex]

1)

S=r θ

S=11([tex]\frac{165\pi }{180}[/tex])........................................plug in values

S=11( [tex]\frac{11\pi }{12}[/tex]).........................................simplify

S=31.68 yd....................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]11^{2}[/tex]) [tex]\frac{165\pi }{180}[/tex]....................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]121[/tex])  [tex]\frac{11\pi }{12}[/tex]....................................simplify

A=174.23 [tex]yd^{2}[/tex].....................................solve and round.

2)

S=r θ

S=14([tex]\frac{150\pi }{180}[/tex])...........................................plug in values

S=  =14(  [tex]\frac{5\pi }{6}[/tex]).........................................simplify

S=36.65 in........................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]14^{2}[/tex])  [tex]\frac{150\pi }{180}[/tex]......................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]196[/tex])  [tex]\frac{5\pi }{6}[/tex]......................................simplify

A=256.56 [tex]in^{2}[/tex]....................................solve and round.

3)

S=r θ

S=3([tex]\frac{170\pi }{180}[/tex]).........................................plug in values

S=  3(  [tex]\frac{17\pi }{18}[/tex]).......................................simplify

S=8.9 ft..........................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]3^{2}[/tex]) [tex]\frac{170\pi }{180}[/tex]........................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]9[/tex])  [tex]\frac{17\pi }{18}[/tex].......................................simplify

A=13.35 [tex]ft^{2}[/tex]........................................solve and round.

4)

S=r θ

S=6([tex]\frac{165\pi }{180}[/tex]).......................................plug in values

S=6( [tex]\frac{11\pi }{12}[/tex]).......................................simplify

S=17.28 in....................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]6^{2}[/tex]) [tex]\frac{165\pi }{180}[/tex].....................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]36[/tex])  [tex]\frac{11\pi }{12}[/tex]..................................simplify

A=51.84 [tex]in^{2}[/tex].....................................solve and round.

5)

S=r θ

S=10([tex]\frac{145\pi }{180}[/tex]).........................................plug in values

S=10(  [tex]\frac{29\pi }{36}[/tex]).........................................simplify

S=25.31 yd.......................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]10^{2}[/tex]) [tex]\frac{145\pi }{180}[/tex].......................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]100[/tex])  [tex]\frac{29\pi }{36}[/tex]....................................simplify

A=126.54 [tex]yd^{2}[/tex].....................................solve and round.

6)

S=r θ

S=3([tex]\frac{65\pi }{180}[/tex])..........................................plug in values

S=  3(  [tex]\frac{13\pi }{36}[/tex]).......................................simplify

S=3.4ft............................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]3^{2}[/tex]) [tex]\frac{65\pi }{180}[/tex].......................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]9[/tex])[tex]\frac{13\pi }{36}[/tex]......................................simplify

A=5.11 [tex]ft^{2}[/tex].........................................solve and round.