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Minchanka [31]
3 years ago
13

A rectangular prism and a rectangular pyramid have congruent bases and heights. The prism has a volume of 900 cubic units. One b

ase edge is 5 units and the height of the prism is 10 units. What are the volume and the dimensions of the base of the pyramid? Show your work.
Mathematics
2 answers:
gtnhenbr [62]3 years ago
6 0

Answer:

The question to your answer is:            V = 300 u³

Step-by-step explanation:

Rectangular prism        900 u³   base = 5 u  height = 10 u

Rectangular pyramid  = ?

Rectangular prism = V = large x height x width

                                 900 = large x 10 x 5

                                 900 = large x 50

                                 900 / 50 = large

                                18 = large

Rectangular pyramid = V = 1/3 L x w x h

                                     V = (1/3) 18 x 5 x 10

                                     V = 300 u³

Anika [276]3 years ago
3 0

Answer:

volume= 300u3

Step-by-step explanation:

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Marysya12 [62]

Answer:

either 20.25π or around 63.6173

Step-by-step explanation:

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What is the factored form of x^12y^18+1
Black_prince [1.1K]

Answer:  The required factored form of the given expression is

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Step-by-step explanation:  We are given to find the factored form of the following algebraic expression:

E=x^{12}y^{18}+1.

We will be using the following formula:

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Now, we have

E\\\\=x^{12}y^{18}+1\\\\=(x^4y^6)^3+1^3\\\\=(x^4y^6+1)\{(x^4y^6)^2-x^4y^6\times1+1^2\}\\\\=(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

Thus, the required factored form of the given expression is

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4 0
3 years ago
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What is the base 5 representation of the number 465^8
Alika [10]

We need to convert given value 465^8 as the base 5 number.

Let us assume x is the exponent of 5 when we take base 5.

We can setup an equation now.

5^x=465^8

Taking ln on both sides, we get

\ln \left(5^x\right)=\ln \left(465^8\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

x\ln \left(5\right)=8\ln \left(465\right)

\mathrm{Divide\:both\:sides\:by\:}\ln \left(5\right)

\frac{x\ln \left(5\right)}{\ln \left(5\right)}=\frac{8\ln \left(465\right)}{\ln \left(5\right)}

On simplifying, we get

x=\frac{8\ln \left(465\right)}{\ln \left(5\right)}

Therefore, 465^8 \ can \ be \ written \ as \ 5^{\frac{8\ln \left(465\right)}{\ln \left(5\right)}}..

3 0
4 years ago
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Evaluate the following integrals: 1. Z x 4 ln x dx 2. Z arcsin y dy 3. Z e −θ cos(3θ) dθ 4. Z 1 0 x 3 √ 4 + x 2 dx 5. Z π/8 0 co
Zigmanuir [339]

Answer:

The integrals was calculated.

Step-by-step explanation:

We calculate integrals, and we get:

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2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}

3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}

4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}

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6)  ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x

7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}

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s344n2d4d5 [400]

Answer:

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Step-by-step explanation:

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The maximum height occurs at:

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We substitute t=2.125 to get:

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4 0
3 years ago
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