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madam [21]
3 years ago
14

Calculate the distance between the points J=(1,-2) and G=(9,-7) and in the coordinate plane. Give an exact answer (not a decimal

approximation). I WILL GIVE 20) POINTS
Mathematics
1 answer:
ira [324]3 years ago
3 0

|JG| = |(9, -7) -(1, -2)|

= |(9 -1, -7 -(-2))|

= |(8, -5)|

= √(8² +(-5)²)

= √(64 +25)

= √89

The distance between J and G is √89.

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Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of k, where k
tresset_1 [31]

Answer:

a) Similar

b) Perimeter of rectangle 2 is k times the Perimeter of rectangle 1 (Proved Below)

c) Area of rectangle 2 is k^2 times the Area of rectangle 1 (Proved Below)

Step-by-step explanation:

Given:

Rectangle 1 has length = x

Rectangle 1 has width = y

Rectangle 2 has length = kx

Rectangle 2 has width = ky

(a) Are Rectangle 1 and Rectangle 2 similar? Why or why not?

Rectangle 1 and Rectangle 2 are similar because the angles of both rectangles are 90° and the sides of Rectangle 2 is k times the sides of Rectangle 1. So sides of both rectangles is equal to the ratio k.

(b) Write a paragraph proof to show that the perimeter of Rectangle 2 is k times the perimeter of Rectangle 1.

Perimeter of Rectangle = 2*(Length + Width)

Perimeter of Rectangle 1 = 2*(x+y) = 2x+2y

Perimeter of Rectangle 2 = 2*(kx+ky) = 2kx + 2ky

                                          = k(2x+2y)

                                          = k(Perimeter of Rectangle 1)

Hence proved that Perimeter of rectangle 2 is k times the perimeter of rectangle 1.

(c) Write a paragraph proof to show that the area of Rectangle 2 is k^2 times the area of Rectangle 1.

Area of Rectangle = Length * width

Area of Rectangle 1 = x * y

Area of Rectangle 2 = kx*ky

                                  = k^2 (xy)

                                  = k^2 (Area of rectangle 1)

Hence proved that area of rectangle 2 is k^2 times the area of rectangle 1.

4 0
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Nat2105 [25]
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