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3 years ago

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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sveta [45]

Answer:

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Step-by-step explanation:

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3 years ago

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 man

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Answer:

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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3 years ago

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gregori [183]

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3 years ago

6. 0, 16, 32, 48, 64, ______, ______, ______ <br> what is the next 3 numbers.

Damm [24]

Answer:

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Step-by-step explanation:

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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