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irakobra [83]
3 years ago
10

Anyone know anything about 10th grade tech geometry? message me please for details.

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
7 0

Answer:

Simply put, geometry is a branch of mathematics that studies the size, shape, and position of 2-dimensional shapes and 3-dimensional figures

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After learning how to fly, Wendy reduced her daily commute time by 75\%75%75, percent. Previously, her commute took mmm minutes.
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Answer: 1/4 and 0.25

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Answer:

AB is A

BC is B

AC is D

Step-by-step explanation:

To find the length of each side, use the formula for the distance between coordinate pairs.

We can find the distance using the distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

<u>AB</u>

We then substitute (-3,6) as (x_1,y_1) and (2,1) as (x_2,y_2).

d=\sqrt{(2--3)^2+(1-6)^2} \\d=\sqrt{(2+3)^2+(-5)^2} \\d=\sqrt{25+25}\\d=\sqrt{50}

<u>BC</u>

We then substitute (2,1) as (x_1,y_1) and (9,5) as (x_2,y_2).

d=\sqrt{(9-2)^2+(5-1)^2} \\d=\sqrt{(-7)^2+(4)^2} \\d=\sqrt{49+16}\\d=\sqrt{65}

<u>AC</u>

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d=\sqrt{(9--3)^2+(5-6)^2} \\d=\sqrt{(12)^2+(-1)^2} \\d=\sqrt{144+1}\\d=\sqrt{145}

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Let $\mu$ and $\sigma^2$ denote the mean and variance of the random variable x. determine $e[(x-\mu)/\sigma]$ and $e{[((x-\mu)/\
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\mathbb E\left(\dfrac{X-\mu}{\sigma}\right)=\dfrac1\sigma\mathbb E(X)-\dfrac\mu\sigma=\dfrac{\mu-\mu}\sigma=0

\mathbb E\bigg(\left(\dfrac{X-\mu}\sigma\right)^2\bigg)=\dfrac1{\sigma^2}\mathbb E\left(X^2-2\mu X+\mu^2\right)=\dfrac{\mathbb E(X^2)-2\mu\mathbb E(X)+\mu^2}{\sigma^2}
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=\dfrac{\mathbb V(X)}{\sigma^2}=\dfrac{\sigma^2}{\sigma^2}=1
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4 years ago
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One simple question.... I see no zeros :
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3 years ago
Factor 3x^2 + 2x -18x - 12
soldier1979 [14.2K]
3x^2+ 2x -18x -12
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= (3x+2)(x-6)

Final answer: (3x+2)(x-6)
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3 years ago
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