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Paraphin [41]
3 years ago
6

1 half dollar + 1 quarter - 2 dimes

Mathematics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

55 cents.

Step-by-step explanation:

First, we start off with a half dollar (50 cents), and add a quarter (25 cents). 50 + 25 = 75. Next we take two dimes (each dime being 10 cents). 10 + 10 is 20, so we have twenty cents in dimes. It says to subtract the 2 dimes from our current number 75 minus 20 is 55. Therefore, we have 55 cents left. (If needed, that is 1 half dollar and 1 nickel.)

Hope this helps! :D

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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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