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zhannawk [14.2K]
3 years ago
8

Trapezoid TRAP was dilated by a scale factor of one half to create trapezoid T'R'A'P'.

Mathematics
2 answers:
Crazy boy [7]3 years ago
8 0

Answer:

X

Step-by-step explanation:

alexira [117]3 years ago
6 0

Answer:

Point X

Step-by-step explanation:

If you look, you can see that X is the center point for both TRAP and T'R'A'P'. I also just took the test myself and got it correct.

Hope this helps!

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The perimeter of the square is 4(x + 3.3) meters. What is the value of x if the perimeter is 28 meters?
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4(x+3.3)=28
/4             /4
x+3.3=7
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0.4r-1.2=0.3r+0.6 can you help me out on this question??
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The answer is R=18 I hope this helps
5 0
4 years ago
NEED HELP! WILL MARK BRAINLIEST
ICE Princess25 [194]

Answer:

im pretty sure its 27.

Step-by-step explanation:

if g(x) = 20 then the new problem is 20=x-7

20=x-7 . add 7 to both sides

+7=+7 . +7 cancels out the -7

27=x . <<<answer

6 0
3 years ago
1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest
Sophie [7]

These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.

<h2>Exercise 1</h2>

You can call four consecutive integers as

x,\ x+1,\ x+2,\ x+3

So, the sum of the first and third is x+(x+2) = 2x+2

We want this quantity to be six less than the largest, i.e. (x+3)-6 = x-3

So, the equality is

2x+2 = x-3

Subtract x from both sides:

x+2 = -3

Subtract 2 from both sides:

x = -5

So, the consecutive integers are

-5,\ -4,\ -3,\ -2

In fact, the sum of the first and third is -5-3 = -8, which is indeed six less than the largest: -8 = -2-6

<h2>Exercise 2</h2>

If you call the first odd number x, the next consecutive odd numbers will be x,\ x+2,\ x+4,\ x+6

In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)

<h2>Exercise 3</h2>

By the same logic of exercise 2, two consecutive even integers are x,\ x+2, assuming that x is even.

So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)

<h2>Exercise 4 to 8</h2>

These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:

x,\ x+1,\ x+2\quad\text{or}\quad  x,\ x+1,\ x+2\ x+3

and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: 3(x+1), while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting: x+6 or x-13.

<h2>Exercise 9</h2>

A multiple of 5 can be written as 5k, for some integer k.

So, three consecutive multiples of 5 are

5k, 5(k+1), 5(k+2) = 5k, 5k+5, 5k+10

We want these three numbers to have a sum of 75. So, we have

5k, 5k+5, 5k+10 = 75 \iff 15k+15 = 75 \iff 15k = 60 \iff k = 4

So, the three numbers are

5k, 5(k+1), 5(k+2) = 5\cdot 4, 5\cdot 5, 5\cdot 6 = 20, 25, 30

3 0
3 years ago
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