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3 years ago

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Mathematics

1 answer:

stiv31 [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Hello There!

The y value of ABC was subtracted by 7 to get to the figure A'B'C'

So the rule for the given transformation would be

(x,y) ---> (x,y-7)

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The sum of the first five terms of an arithmetic series is 25, and the first term is 2. Find the constant difference.

aivan3 [116]

Step-by-step explanation:

if you calculate both of you should ask her

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3 years ago

Area = 120 units squared <br> Base = 8 units <br> Length = ? (Include the units in your answer)

Vesnalui [34]

Answer:

Step-by-step explanation:

Okay, so we know that Base*Length=Area, so we just have to divide the area by the base, so the formula here is Area/Base=Length.

That would get us 120/8=Length,

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2 years ago

The rectangular box is used to ship mp3 players to an electronics store

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Answer:

wasuppp

Step-by-step explanation:

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3 years ago

Complete the square to re-write the quadratic function in vertex form: y = x ^ 2 - 2x - 8

blondinia [14]

Answer:

(1,-9)

Step-by-step explanation:

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(h,k).

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2 years ago

Read 2 more answers

PLEASE HELP ME IVE POSTED THIS 765678 AND STILL NO RESPONSE

Debora [2.8K]

Problem 1

<h3>Answer: 6.7</h3>

----------------

Work Shown:

The two points are $(x_1,y_1) = (1,-2)$ and $(x_2,y_2) = (4,4)$

Apply the distance formula to get the following

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (-2-4)^2}\\\\d = \sqrt{(-3)^2 + (-6)^2}\\\\d = \sqrt{9 + 36}\\\\d = \sqrt{45}\\\\d \approx 6.7082039\\\\d \approx 6.7\\\\$

The distance between the two endpoints is roughly 6.7 units. This is the same as saying the segment is roughly 6.7 units long.

======================================================

Problem 2

<h3>Answer: 3.6</h3>

----------------

Work Shown:

We'll use the distance formula here as well.

This time we have the two points $(x_1,y_1) = (3,1)$ and $(x_2,y_2) = (5,-2)$

The distance between them is...

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-5)^2 + (1-(-2))^2}\\\\d = \sqrt{(3-5)^2 + (1+2)^2}\\\\d = \sqrt{(-2)^2 + (3)^2}\\\\d = \sqrt{4 + 9}\\\\d = \sqrt{13}\\\\d \approx 3.6055513\\\\d \approx 3.6\\\\$

This distance is approximate.

5 0

3 years ago