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3 years ago

(1.08 × 10−3)(9.3 × 10−3)

Mathematics

2 answers:

Vera_Pavlovna [14]3 years ago

7 0

Answer:

2509⋅(3x 5+5)⋅(3x 5 −5)⋅(31x 10 −10)

Step-by-step explanation:

ElenaW [278]3 years ago

4 0

Answer:

694.2

Step-by-step explanation:

first you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^

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Jake plans to make and sell wooden birdhouses. He plans to sell each birdhouse for $50, his income can be found using the equati

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3 years ago

A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h,

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Answer:

7.41

Step-by-step explanation:

Vertical component:

S = 268.8

U = 0

V =?

A = 9.8

T=?

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3 years ago

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Identify the construction shown in the diagram.

VMariaS [17]

<h3>Answer: Choice C) Construct a tangent from an external point</h3>

=================================

Explanation:

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3 years ago

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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