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2 years ago

AABC has been translated 5 units to the right, as shown in the diagram. What

Mathematics

1 answer:

ANEK [815]2 years ago

4 0

The length of ab is D. 15

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Debra has ridden 57 miles of a bike course the course is 60 miles long what percentage of the course has Debra ridden so far

Reptile [31]

Hello!

To find our answer, we just divide.

57/60=0.95

We multiply by 100 to find our percent

100(0.95)-95

Therefore, our answer is 95%

I hope this helps!

4 0

3 years ago

Read 2 more answers

Find the equation of the line. Use exact numbers. Y

Andrews [41]

Answer:

Y=-1/4X-6

Step-by-step explanation:

y=--1/4X-6

$y = \frac{ - 1}{4} x - 6$

7 0

2 years ago

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The sum of three numbers is 102. The first number is 6 more than the second. The third number is 4 times the second. What are th

const2013 [10]

Answer:

1st number: 22

2nd number: 16

3rd number: 64

Step-by-step explanation:

x + y + z = 102

x = 6 + y

z = 4y

plug that in!

(6 + y) + y + (4y) = 102

get rid of the parenthesis (I added them so you could see what I was replacing) and add like terms.

6 + 6y = 102

move 6 to the other side.

6y = 96

y = 16

now that you know one number, you can solve the equations for the rest!

x = 6 + (16)

z = 4(16)

__________________________________________________________

x = 22

y = 16

z = 64

6 0

2 years ago

Simplify (3y^6)^5/(y^3)^5

Alex17521 [72]

Answer: 245y^15

Step-by-step explanation:

5 0

2 years ago

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What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?

natulia [17]

Answer:

The endpoint coordinates for the mid-segment are: $(-2,-1)$ and $(1,0)$

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of $\triangle BCD$ are: $B(-3,1), C(3,3)$ and $D(-1,-3)$

Now, the endpoints of the mid-segment of $\triangle BCD$ which is parallel to $BC$ will be the mid-points of sides $BD$ and $CD$ .

Formula for the coordinate of mid-point : $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two endpoints.

So, the mid-point of $BD$ will be: $(\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)$

and the mid-point of $CD$ will be: $(\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)$

Thus, the endpoint coordinates for the mid-segment of $\triangle BCD$ that is parallel to $BC$ are $(-2,-1)$ and $(1,0)$

4 0

2 years ago

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