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erma4kov [3.2K]

3 years ago

6

How do I factorize the following: 3x+33 x-33 -3x+33 30x cubed-5x squared

1 answer:

erica [24]3 years ago

8 0

By favoriting the following

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3. Bricks are delivered to a work site and stacked in rows and columns, forming a rectangular prism. The length of the prism is

pychu [463]

Volume of a rectangular prism = 40 cubic feet.

Volume of a rectangular prism = length * width * height

width = x
length = 1 foot greater than its width = x + 1
height = 2 feet less than its width = x - 2

l * w * h = 40 ft³
(x+1) * x * (x -2) = 40
(x² + x)(x-2)
x³ - 2x² + x² + 2x = 40
x³ - x² + 2x - 40 = 0

solve for the value of x. I used the calculator to get the volume.

width = x = 4
length = x + 1 = 4 + 1 = 5
height = x - 2 = 4 - 2 = 2

V = 4 * 5 * 2
V = 20 * 2
V = 40 ft³

3 0

3 years ago

Germaine ate 1/4 bag of popcorn that was 1/2 full.

valentina_108 [34]

Simply multiply the fractions.

1/4 * 1/2 = 1/8 Choice A.

1/8 of the whole bag of popcorn is the amount of popcorn Germaine ate.

8 0

3 years ago

Read 2 more answers

What must an animal do in order for cellular respiration to begin

Wittaler [7]

Answer:

An animal must consume food, inhale oxygen, and drink water to begin cellular respiration

Step-by-step explanation:

7 0

3 years ago

Using the identity sin²theta + cos²theta = 1, find the value of tan , to the nearest hundredth, if cos theta = 0.57 and 0 < t

n200080 [17]

Answer:

tan*=sin*/cos*

sin*=1-(0.57)^2

sin*=0.6751

tan*=0.57/0.6751

=0.844

8 0

3 years ago

What expression is equivalent to ^3 square root of 2y^3 times 7 square root of 18y

Mumz [18]

Answer:

$\sqrt[3]{2y^3} * 7\sqrt{18y} = 21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

Step-by-step explanation:

The question is poorly formatted.

Given

$\sqrt[3]{2y^3} * 7\sqrt{18y}$

Required

Derive an equivalent expression

$\sqrt[3]{2y^3} * 7\sqrt{18y}$

Express 18 as 9 * 2

$\sqrt[3]{2y^3} * 7\sqrt{9 * 2y}$

Split the expression as follows:

$\sqrt[3]{2y^3} * 7\sqrt{9} * \sqrt{2y}$

Take positive square root of 9

$\sqrt[3]{2y^3} * 7*3 * \sqrt{2y}$

$\sqrt[3]{2y^3} * 21 * \sqrt{2y}$

$21*\sqrt[3]{2y^3} * \sqrt{2y}$

The cube root can be rewritten to give:

$21*\sqrt[3]{2}*\sqrt[3]{y^3} * \sqrt{2y}$

$\sqrt[3]{y^3} = y^{3*\frac{1}{3}} = y$

So, we have:

$21*\sqrt[3]{2} * y * \sqrt{2y}$

Rewrite as:

$21y *\sqrt[3]{2} * \sqrt{2y}$

Split $\sqrt{2y$

$21y *\sqrt[3]{2} * \sqrt{2} * \sqrt{y}$

Collect Like Terms

$21y*\sqrt{y} *\sqrt[3]{2} * \sqrt{2}$

Represent in index form

$21y*y^{\frac{1}{2}} *2^\frac{1}{3} *2^\frac{1}{2}$

Apply law of indices

$21*y^{1+\frac{1}{2}} *2^{\frac{1}{3} +\frac{1}{2} }$

$21*y^{\frac{2+1}{2}} *2^{\frac{2+3}{6}}$

$21*y^{\frac{3}{2}} *2^{\frac{5}{6}}$

$21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

Hence:

$\sqrt[3]{2y^3} * 7\sqrt{18y} = 21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

6 0

3 years ago