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tester [92]
3 years ago
8

PLEASE HELP DUE TODAY!!

Mathematics
1 answer:
telo118 [61]3 years ago
7 0

9514 1404 393

Answer:

  2.8¢, true

Step-by-step explanation:

The value of copper in a penny is reportedly about ...

  (0.95 × 3.11 g)×(1 oz)/(28.35 g)×($0.27/oz) ≈ $0.028

Today, the value of the copper in a pre-1982 penny is about 2.8¢, more than double the value of the penny as legal tender.

The claim is true.

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The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
Arada [10]

Given:

Scale factor s=\dfrac{1}{3}

Center of dilation = (4,2)

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The coordinates of the points C' and A.

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We know that, 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)

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)

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

Using rule (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)

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

Let us assume that point A is A(m,n).

Using rule (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).

4 0
2 years ago
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