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3 years ago

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Look at this prism and its net. Rectangular prism with length labeled 5 meters, width labeled 7 meters, and height labeled 12 me

ters. Net of the same rectangular prism with the same dimensionsn labeled. What is the surface area of this rectangular prism? cm²

1 answer:

lord [1]3 years ago

4 0

The surface area is <span>358.</span>

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PLZ HELP SOLVE: -4 < k + 3 < 8

vaieri [72.5K]

Hope this was what you’re looking for ^.^

(If correct, if you don’t mind can you please mark me as the brainliest, it’s okay if not)

3 0

4 years ago

Simplifying rational expressions <br> x^2-16/x^2+6x+8 = x-4/x+2<br> x^2-x-6/x^2-3x-10 = x-3/x-5

enot [183]

1)

$\dfrac{x^2-16}{x^2+6x+8}$

Decompose the numerator and denominator into multipliers

To simplify the numerator we use the formula of difference of squares

$x^2-y^2=(x-y)(x+y)$

$x^2-16=(x-4)(x+4)$

To decompose the denominator into multipliers solve the square equation

$x^2+6x+8=0\\D=6^2-4*8=4=2^2\\x_1=\dfrac{-6+2}{2} =-2\\x_2=\dfrac{-6-2}{2} =-4$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-(-2))(x-(-4))=(x+2)(x+4)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$ , so

$\dfrac{x^2-16}{x^2+6x+8}=\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$

2)

$\dfrac{x^2-x-6}{x^2-3x-10}$

Decompose the numerator and denominator into multipliers

To decompose the numerator into multipliers solve the square equation

$x^2-x-6=0\\D=(-1)^2-4*(-6)=25=5^2\\x_1=\dfrac{1+5}{2} =3\\x_2=\dfrac{1-5}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-3)(x-(-2))=(x-3)(x+2)$

To decompose the denominator into multipliers solve the square equation

$x^2-3x-10=0\\D=(-3)^2-4*(-10)=49=7^2\\x_1=\dfrac{3+7}{2} =5\\x_2=\dfrac{3-7}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-5)(x-(-(-2))=(x-5)(x+2)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$ , so

$\dfrac{x^2-x-6}{x^2-3x-10}=\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$

Hello from Russia:^)

6 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>

7 0

4 years ago

Help please I need this immediately thanks.

Nezavi [6.7K]

Answer:

He did not make a mistake

3 0

3 years ago

Read 2 more answers

What is the volume of a sphere with a diameter of 7.7 ft, rounded to the nearest tenth

Scorpion4ik [409]

Step-by-step explanation:

V=4/3πr^3

V=4/3π(3.85)^3

V=4/3π(57.066625)

V=4/3 (179.280089865)

V=239.04011982

V=239 ft^3

8 0

3 years ago