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3 years ago

Which integer is closest to 0 on the number line?

Mathematics

2 answers:

marishachu [46]3 years ago

7 0

1 is the closest to 0 on the number line. Hope this helps!!!!!!!!

HACTEHA [7]3 years ago

5 0

C. 1 is closest to 0 on the number line because it is only one number away from 0 which is the closest distance.

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3 years ago

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2 years ago

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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

3 years ago

Prove algebraically that r = 10/2+2sinTheta is a parabola

Xelga [282]

Answer:

$y = - \frac{ 1 }{10} {x}^{2} + \frac{5}{2}$

Step-by-step explanation:

We want to prove algebraically that:

$r = \frac{10}{2 + 2 \sin \theta}$

is a parabola.

We use the relations

${r}^{2} = {x}^{2} + {y}^{2}$

and

$y = r \sin \theta$

Before we substitute, let us rewrite the equation to get:

$r(2 + 2 \sin \theta) = 10$

$r(1+ \sin \theta) = 5$

Expand :

$r+ r\sin \theta= 5$

We now substitute to get:

$\sqrt{ {x}^{2} + {y}^{2} } + y = 5$

This means that:

$\sqrt{ {x}^{2} + {y}^{2} }=5 - y$

Square:

${x}^{2} + {y}^{2} =(5 - y)^{2}$

Expand:

${x}^{2} + {y}^{2} =25 - 10y + {y}^{2}$

${x}^{2} =25 - 10y$

${x}^{2} - 25 = - 10y$

$y = - \frac{ {x}^{2} }{10} + \frac{5}{2}$

This is a parabola (0,2.5) and turns upside down.

4 0

3 years ago