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Citrus2011 [14]
3 years ago
14

I really need to know the answer this fast please!!!!!!

Mathematics
1 answer:
Zepler [3.9K]3 years ago
3 0
The answer is A I think
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The sum of two numbers is -5. The difference of the two number is -17. What is the product of the two numbers?
natulia [17]
The sum is 20 okay by the way it’s wrong u
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3 years ago
What is the value of z in the diagram?
lutik1710 [3]

Answer:

2

Step-by-step explanation:

4 0
3 years ago
Trevor picked 5 ⅗ pounds of cherries. He distributed them evenly into 12 containers. How many pounds of cherries were in each co
natulia [17]

Answer:

7/15

Step-by-step explanation:

For me, it is easier to take the 3/5 and put it in decimal form which is .6 and add the 5 to it, making 5.6, and then in a calculator put in 5.6 divided by 12 and you should get a big decimal that's pretty big but if you have a graphing calculator and convert that to a fraction and you should get 7/15, This is a pretty lazy way of me doing it but that's how I did it.

3 0
3 years ago
What is the image of (-9,6) after a dilation by a scale factor of 4 centered at the<br> origin?
myrzilka [38]

Answer:

<em>( - 36, 24 ) </em>

Step-by-step explanation:

(- 9, 6) ------> ( - 9×4, 6×4) = <em>( - 36, 24 )</em>

3 0
2 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

5 0
2 years ago
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