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tangare [24]
3 years ago
5

A right rectangular prism has a square base with side lengths of 5 inches and a height of 9 inches. What is the volume of the ri

ght rectangular prism?
A.
90 cubic inches
B.
225 cubic inches
C.
19 cubic inches
D.
45 cubic inches
Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
4 0

Answer:

  B.  225 cubic inches

Step-by-step explanation:

The volume is the product of the base area and the height:

  V = Bh

The area of the square base is the product of the side lengths:

  B = (5 in)^2 = 25 in^2

For a height of 9 in, the volume is ...

  V = (25 in^2)(9 in) = 225 in^3

The volume of the right rectangular prism is 225 cubic inches.

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Answer:

  • (x² + 2y² + 2xy)(x² + 2y² - 2xy)

Step-by-step explanation:

<h3>Given binomial:</h3>

x⁴ + 4y⁴

In order to factorize it, follow the steps:

<h3>Complete the square:</h3>

  • x⁴ + 4y⁴ =
  • (x²)² + 2*(x²)*(2y²) + (2y²)² - 2*(x²)*(2y²) =
  • (x² + 2y²)² - (2xy)² =
  • (x² + 2y² + 2xy)(x² + 2y² - 2xy)

Used identities:

  • (a + b)² = a² + 2ab + b²
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4 0
2 years ago
Please answer... I Got Pointsss
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Answer:

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Step-by-step explanation:

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7 0
4 years ago
A document that is 35 in. By 40 in. Is redrawn at 1 1/2 scale and then redrawn at 1/4 scale. What are the final dimensions of th
Andru [333]
The dimension of the document is 35 inches by 40 inches.
It is redrawn at a scale of 1 1/2 or 3/2 or 1.5
The dimension will be:
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Then redrawn again at 1/4 or 0.25
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3 years ago
Which set of measures can represent the lengths of a right triangle?
Ivanshal [37]

Answer:

A and B are correct, but A is most correct.

Step-by-step explanation:

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4 0
3 years ago
Find the limit
Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

\large\underline{\sf{Solution-}}

Given expression is

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

On substituting directly x = 1, we get,

\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}

\rm \: = \sf \: \: - \infty \: - \: \infty

which is indeterminant form.

Consider again,

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

can be rewritten as

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]

\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}

\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}

\rm \: = \: \sf \boxed{2}

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}

\rule{190pt}{2pt}

7 0
3 years ago
Read 2 more answers
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