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3 years ago

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PLEASE HELP!!!!!

1 answer:

Sidana [21]3 years ago

3 0

Answer:

Part a) The distance on a map between Joseph's house and the airport is 2.53 inches

Part b) The distance on a map between the airport and the restaurant is 1.68 inches and the total distance on a map between Joseph's house and the restaurant is 4.21 inches

Step-by-step explanation:

Part a) The actual distance between Joseph's house and the airport is 24 miles. How far apart are Joseph's house and the airport on the map?

we know that

The scale of a map is 1 inch : 9.5 miles

so

using proportion

Find the distance on a map if the actual distance between Joseph's house and the airport is 24 miles

Let

x-----> the distance on a map

1/9.5=x/24

x=24/9.5=2.53 inches

Part b) Joseph traveled from his house to the airport. He then traveled another 16 miles past the airport to a restaurant. How many inches on the map represent this distance?

we know that

The scale of a map is 1 inch : 9.5 miles

so

using proportion

Find the distance on a map if the actual distance between airport to the restaurant is 16 miles

Let

x-----> the distance on a map

1/9.5=x/16

x=16/9.5=1.68 inches

The total distance on a map between Joseph's house and the restaurant is equal to

2.53 inches+1.68 inches=4.21 inches

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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

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Given:

The red figure dilated with a scale factor of $s=\dfrac{1}{3}$ and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

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

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

In given problem, 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 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).

4 0

3 years ago