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Kipish [7]
3 years ago
8

Find the volume of this rectangular prism by using the formula V=Bh.

Mathematics
2 answers:
Stolb23 [73]3 years ago
7 0

Answer:

Step-by-step explanation:

The volume of a rectangular prism is

V=xyz, where x,y,z are the dimensions. (I don’t know where you got the faulty V=Bh)

V=3.2(5)9

V=144in^3

AVprozaik [17]3 years ago
5 0

Answer:

Find the volume of the rectangular prism using the formula V = Bh.

The area of the base, B, equals

✔ 21

cm2.

The volume of the prism is

✔ 157 1/2

cm3.

Step-by-step explanation:

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1 / 2 or 0.5

Step-by-step explanation:

use the formula ; m = y2 - y1 / x2 - x1

where y2 = 10

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Step-by-step explanation:

A printer takes two and one half seconds to print a flyer it took 75 seconds to print a batch of flyers without stopping how many flyers were printed with a multiplication and division equation and solve?

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one thousand tickets are sold at $2 each. One ticket will be selected and the winner will receive a flat screen TV valued at $38
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8 0
3 years ago
6. 0, 16, 32, 48, 64, ______, ______, ______ <br> what is the next 3 numbers.
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Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
Let <img src="https://tex.z-dn.net/?f=%5Calpha" id="TexFormula1" title="\alpha" alt="\alpha" align="absmiddle" class="latex-form
Margarita [4]

Answer:

|\alpha| = 2

Step-by-step explanation:

Since \alpha and \beta are complex conjugates, let's define them as follows:

\alpha = a+bi

\beta = a-bi

\frac{\alpha}{\beta^2}=\frac{a+bi}{a^2-b^2-2abi} =\frac{(a+bi)*(a^2-b^2+2abi)}{(a^2-b^2-2abi)*(a^2-b^2+2abi)} =\frac{a^3-3ab^2+(3a^2b-b^3)i}{a^4+2a^2b^2+b^4}

Since \frac{\alpha}{\beta^2} is a real number, complex part of above result must be zero.

3a^2b-b^3=0

From to hold above equality, b=0 or b^2=3a^2.

However, since |\alpha-\beta|=2\sqrt 3, b\neq 0

So, b =\sqrt 3 a or b =-\sqrt 3 a

And since \alpha and \beta are complex conjugates, taking plus or minus sign as found above will not affect the result, so let's write the last version of \alpha and \beta as follows:

\alpha = a+\sqrt 3 ai

\beta = a-\sqrt 3 ai

Since |\alpha-\beta|=2\sqrt 3

|a+\sqrt 3 ai-a+\sqrt 3 ai|=2\sqrt 3 ⇒ a = 1

Finally, \alpha = 1+\sqrt 3 i ⇒ |\alpha| = \sqrt{(1^2+\sqrt 3^2)}=2

3 0
3 years ago
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