What is the value of x? x20=58 A. 0.625 B. 2.5 C. 12.5 D. 32.0

One of Mr. Grayson’s customers, Zenani, is creating an art wall on which she will display some art in triangular frames. She wan

Vlada [557]

A) There are two possible measurements for the 3rd side. It depends of if you want the 16in piece to be a leg of the hypotenuse. If the 16in side is a leg, then the 3rd side would measure 20.6in. If the 16in side is the hypotenuse, then the 3rd side would measure 9.3in.
b) Either of these triangles could be doubled to make the 2nd triangle. You would have sides of 26in, 32in, and 41.2in. Or you would have sides measuring 18.6in, 26in, and 32in.
c) Because the sides of a right triangle have to satisfy the Pythagorean Theorem adding the same number to all three sides will not create a right triangle. If we are letting the 16in side be a leg
(13+x)²+(16+x)²=(20.6+x)² would have to be true. If you multiply it out and combine terms you find that. 2x²+58x+425=x²+41.2x+425. If we try to solve for x we get x²+16.8x=0. This has two solutions: 0 and -16.8. Neither of those work in this situation because adding zero wouldn't change the triangle and adding the negative would leave you with sides of negative length. You would need up with similar results if you tried it with the other set of numbers.

7 0

3 years ago

What is 5/8 written as a decimal

Vera_Pavlovna [14]

What is 5/8 written as a decimal?

5/8 = 5 : 8 = 0.625

8 0

3 years ago

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3. Which measurements would represent the

Mkey [24]

Answer:

B

Step-by-step explanation:

6 0

3 years ago

Prove the diagonals of the square with vertices P (0,4), Q (4,4), R (0,0) and S (4,0) are perpendicular bisectors of each other.

Aneli [31]

Answer: The slope of diagonal PS is -1. The The slope of diagonal QR is 1. The midpoint of PS is (2,2). Midpoint of QR is (2,2).

Explanation:

The given vertices are P (0,4), Q (4,4), R (0,0) and S (4,0).

If a line passing through the point $(x_1,y_1)$ and $(x_2,y_2)$ , then the slope of the line is,

$m=\frac{y_2-y_1}{x_2-x_1}$

The slope of diagonal PS is,

$m_1=\frac{0-4}{4-0}= \frac{-4}{4} =-1$

The slope of diagonal QR is,

$m_2=\frac{0-4}{0-4}= \frac{-4}{-4} =1$

The slope of both diagonals are 1 and -1, therefore the product of their slopes is -1. So, the diagonals are perpendicular to each other.

The formula for midpoint is,

$midpoint=(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )$

The midpoint of PS is,

$\text{midpoint of PS}=(\frac{0+4}{2} ,\frac{4+0}{2})=(2,2)$

The midpoint of QR is,

$\text{midpoint of QR}=(\frac{4+0}{2} ,\frac{4+0}{2})=(2,2)$

The midpoint of both diagonals are same, therefore diagonals are bisecting each other.

Therefore, the slope of diagonal PS is -1. The The slope of diagonal QR is 1. The midpoint of PS is (2,2). Midpoint of QR is (2,2).

5 0

3 years ago

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Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

Answer:

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

Solution:

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0

4 years ago