Draw the line of reflection that reflects △ABC onto triangle Δ A'B'C'

Answer:

48.00

Step-by-step explanation:

Answer:

48.0 hope it helps you out

Answer:

-11

Step-by-step explanation:

[tex]3( - 2) + 5( - 1)[/tex]

[tex] - 6 + ( - 5)[/tex]

[tex] - 11[/tex]

Answer:

D. 8

Step-by-step explanation:

C= 2πr

π= 3.14

C= 25.12

25.12 = 2 * 3.14 * r

r= 4

d= 2r

d= 8

Answer:

think it is d

Step-by-step explanation:

Answer:

8 1/12 boxes

Step-by-step explanation:

Your first step to solve this problem is to convert those fractions to a common denominator. I would choose 12. So your first would be 4 4/12 and your second would be 3 9/12. You would then add those up and get 8 1/12 boxes of muffins bought at the bake sale.

Answer:

60 tiles left over

Step-by-step explanation:

the lobby needs 30 x 8 = 240 tiles.

she has 6 x 50 = 300 tiles

300 - 240 = 60 left over

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Answer: 22.02

Explanation:

I hope this helped!

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Answer: D. Zero

Step-by-step explanation:

It’s a flat line. Flat horizontal lines are ALWAYS zero

Answer:

A

Step-by-step explanation:

(y-5) / (7-5) = (x-1) / (2-1)

(y-5) / 2 = x-1

y-5 = 2x -2

y = 2x +3

Answer:

the answer is X=-7

,,,,,

Answer:

it would be x=7

Step-by-step explanation:

brainliest pls

Answer:

a) 2x² - x + 3 = x(2x + 1) - 2x

<=> 2x² - x + 3 = 2x² + x - 2x

<=> 3 - x = - x

<=> 3 = 0

=> no solution

b) x/2 - x/3 - x/4 = 1/12

[tex] \frac{6x}{12} - \frac{4x}{12} - \frac{3x}{12} = \frac{1}{12} \\ \\ \frac{ - x}{12} = \frac{1}{12} \\ \\ = > - x = 1 \\ \\ \\ < = > x = 1[/tex]

=> the euqation has the solution x = 1

c) |x - 5| = 2|x|

=> x - 5 = 2x

or x - 5 = -2x

<=> x = -5

or 3x = 5

<=> x = -5

x = -5or x = 5/3

d) defined conditions: 2 - x 》 0

<=> x 《 2

we have:

[tex] \sqrt{x + 4} = 2 - x \\ < = > x + 4 = (2 - x) {}^{2} \\ < = > x + 4 = {x}^{2} - 4x + 4 \\ < = > {x}^{2} - 5x = 0 \\ < = > x(x - 5) = 0 \\ [/tex]

<=> x = 0 (because x 《 2)

e)

[tex] \sqrt{5} = \frac{1}{25 {}^{x} } \\ = > 5 = \frac{1}{25 {}^{2x} } \\ < = > 5.25 {}^{2x} = 1 \\ < = > 5.5 {}^{4x} = 1 \\ < = > 5 {}^{4x + 1} = 1 \\ = > 4x + 1 = 0[/tex]

<=> x = -1/4

Answer:

[tex]a) [x: - \infty < x < \infty ][/tex]

[tex]b)[x: - 1][/tex]

[tex]c) [x:- \frac{5}{3} ][/tex]

[tex]d) [x: 3, 0][/tex]

[tex]e) [x: - \frac{1}{4} ][/tex]

Answer:

after the collision cart 1 will stop and cart 2 will keep moving will 0.1.m/s after stop.

Well, the answer would bee 1/10 for a simple reason: the 3 in 7.315 is in the tenth's spot while the 3 in 893.5 is in the ones. a one spot different that can be proven with logic.

Answer:

29/40

Step-by-step explanation:

1/8 + 3/5 =

We need a common denominator.

8 = 2^3

5 = 5

8 and 5 have no common factors, so the LCD is their product.

LCD = 5 * 8 = 40

1/8 + 3/5 =

= 5/40 + 24/40

= 29/40

Answer:

[tex]\boxed {\boxed {\sf A \approx 3.3 \ in^2}}[/tex]

Step-by-step explanation:

Since we are given the central angle in degrees, we should use the following formula for sector area.

[tex]A= \frac {\theta}{360}* \pi r^2[/tex]

The angle is 42 degrees and the radius is 3 inches. Therefore,

[tex]\theta= 42 \\r=3 \ in[/tex]

[tex]A=\frac {42}{360} * \pi (3 \ in)^2[/tex]

Solve the exponent.

[tex]A=\frac {42}{360} * \pi (9 \ in^2)[/tex]

Multiply all three numbers together.

[tex]A= 0.116666666667* 3.14159265359 * 9 \ in^2[/tex]

[tex]A=3.29867228627 \ in^2[/tex]

Let's round to the tenth place. The 9 in the hundredth place tells us to round the 2 up to a 3.

[tex]A \approx 3.3 \ in^2[/tex]

The area of the sector is approximately 3.3 square inches.

Answer:

The number of students that are left handed

Step-by-step explanation:

Answer:

it represent 3 students

Answer:

[tex]P(X \le 4) = 0.7373[/tex]

[tex]P(x \le 15) = 0.0173[/tex]

[tex]P(x > 20) = 0.4207[/tex]

[tex]P(20\ge x \le 24)= 0.6129[/tex]

[tex]P(x = 24) = 0.0236[/tex]

[tex]P(x = 15) = 1.18\%[/tex]

Step-by-step explanation:

Given

[tex]p = 80\% = 0.8[/tex]

The question illustrates binomial distribution and will be solved using:

[tex]P(X = x) = ^nC_xp^x(1 - p)^{n-x}[/tex]

Solving (a):

Given

[tex]n =5[/tex]

Required

[tex]P(X\ge 4)[/tex]

This is calculated using

[tex]P(X \le 4) = P(x = 4) +P(x=5)[/tex]

This gives:

[tex]P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}[/tex]

[tex]P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0[/tex]

[tex]P(X \le 4) = 0.4096 + 0.32768[/tex]

[tex]P(X \le 4) = 0.73728[/tex]

[tex]P(X \le 4) = 0.7373[/tex] --- approximated

Solving (b):

Given

[tex]n =25[/tex]

i)

Required

[tex]P(X\le 15)[/tex]

This is calculated as:

[tex]P(X\le 15) = 1 - P(x>15)[/tex] --- Complement rule

[tex]P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)[/tex]

[tex]P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}[/tex]

[tex]P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0[/tex]  

[tex]P(x > 15) = 0.98266813045[/tex]

So:

[tex]P(X\le 15) = 1 - P(x>15)[/tex]

[tex]P(x \le 15) = 1 - 0.98266813045[/tex]

[tex]P(x \le 15) = 0.01733186955[/tex]

[tex]P(x \le 15) = 0.0173[/tex]

ii)

[tex]P(x>20)[/tex]

This is calculated as:

[tex]P(x>20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)[/tex]

[tex]P(x > 20) = 12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0[/tex]

[tex]P(x > 20) = 0.42067430925[/tex]

[tex]P(x > 20) = 0.4207[/tex]

iii)

[tex]P(20\ge x \le 24)[/tex]

This is calculated as:

[tex]P(20\ge x \le 24) = P(x = 20) + P(x = 21) + P(x = 22) + P(x =23) + P(x = 24)[/tex]

[tex]P(20\ge x \le 24)= 53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1[/tex]

[tex]P(20\ge x \le 24)= 0.61291151859[/tex]

[tex]P(20\ge x \le 24)= 0.6129[/tex]

iv)

[tex]P(x = 24)[/tex]

This is calculated as:

[tex]P(x = 24) = 25* 0.8^{24}*0.2^1[/tex]

[tex]P(x = 24) = 0.0236[/tex]

Solving (c):

[tex]P(x = 15)[/tex]

This is calculated as:

[tex]P(x = 15) = {25}^C_{15} * 0.8^{15} * 0.2^{10}[/tex]

[tex]P(x = 15) = 3268760 * 0.8^{15} * 0.2^{10}[/tex]

[tex]P(x = 15) = 0.01177694905[/tex]

[tex]P(x = 15) = 0.0118[/tex]

Express as percentage

[tex]P(x = 15) = 1.18\%[/tex]

The calculated probability (1.18%) is way less than the advocate's claim.

Hence, we do not believe the claim.

Answer:

hi! how can i help? <3

Step-by-step explanation:

okay i can stand in for you

Answer:

Aaron must obtain a 96 or higher to achieve the desired score to earn an A in the class.

Step-by-step explanation:

Given that the average of Aaron's three test scores must be at least 93 to earn an A in the class, and Aaron scored 89 on the first test and 94 on the second test, to determine what scores can Aaron get on his third test to guarantee an A in the class, knowing that the highest possible score is 100, the following inequality must be written:

93 x 3 = 279

89 + 94 + S = 279

S = 279 - 89 - 94

S = 96

Thus, at a minimum, Aaron must obtain a 96 to achieve the desired score to earn an A in the class.

Aaron must score atleast 90 marks in third test.

Aaron scores in the first  test = 89

Aaron scores in the second test = 94

And the average of Aaron's three test scores must be at least 93 to earn an A.

Let Aaron scores in the third  test = s

Then,

Average = sum of terms divided by the number of terms.

[tex]\frac{89+94+s}{3} \leq 93\\89+94+s\leq 93\times3\\183+s\leq 279\\s\leq 279-189\\s\leq 90[/tex]

So, Aaron must score atleast 90 marks in third test.

Learn more:https://brainly.com/question/19982277

Answer:

im going to guess the last answer. theone that i can't see

Step-by-step explanation:

Answer:

466.08 m²

Step-by-step explanation:

The figure is composed of a triangle and a semicircle.

Therefore,

Area of the figure = area of the triangle + area of the semicircle

= ½*b*h + ½*πr²

Where,

b = 24 m

h = 20 m

π = 3.14

r = ½ of 24 m = 12 m

Substitute each value into the equation

Area of the figure = ½*24*20 + ½*3.14*12²

= 240 + 226.08

= 466.08 m²

Answer:

A the answer is A ^w^ hope this helps

Answer:

The annual percentage rate of change for the population of Oregon is of 8.125%.

Step-by-step explanation:

Total percentage change:

Change multiplied by 100 and divided by the initial value.

Change: 8.7 - 4.8 = 3.9 million

Initial value: 4.8

Percentage change: 3.9*100/4.8 = 81.25%

Annual percentage rate of change

81.25% during 10 years(from 2000 to 2010).

So, per year

81.25%/10 = 8.125%

The annual percentage rate of change for the population of Oregon is of 8.125%.

Answer:

What grade is this 12th?

Step-by-step explanation:

The equivalent expression for the expression given; (( 5/7 )² × (1/3)-³)-¹ is; (7/5)² • (1/3)³.

It follows from the task content that the expression which is equivalent to that which is given is to be determined.

here, we have,

Hence, since the given expression is; (( 5/7 )² × (1/3)-³)-¹;

It follows from the power of powers law of indices that we have;

( 5/7 )-² • (1/3)³

Also, according to the negative exponent law of indices; we have;

(7/5)² • (1/3)³.

On this note, the required equivalent expression is; (7/5)² • (1/3)³.

Read more on equivalent expressions;

brainly.com/question/24734894

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Answer:

139

Step-by-step explanation:

supplementary angles add up to equal 180

so to find the measure of angle ABC we subtract the given angle from 180

180-41=139

so we can conclude that angle ABC=  139

Step-by-step explanation:

< ABC and < CBD are supplementary angles.

< CBD = 41°

< ABC + < CBD = 180° ( Being supplementary angles)

< ABC + 41° = 180°

< ABC = 180° - 41°

< ABC = 139°

Hope it will help :)❤