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2 years ago

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13) Brainliest and 10 + Points!! :) HELP ASAP PLS

1 answer:

Marina CMI [18]2 years ago

4 0

Answer: $2x^2+6x+20=0$

Step-by-step explanation:

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Which of the following represents a valid probability distribution?

elena-14-01-66 [18.8K]

Answer:A

Step-by-step explanation: on edgunity

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2 years ago

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What is the classification for this polynomial ? X^2y + 3x^3

svp [43]

I would say it is considered a binomial

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2 years ago

write an equation of the line with the same slope as the line given by 5x+2y=8 and as the same y-intercept as 3x-7y=10

Paladinen [302]

The slope of the first equation would be -5/2 because
2y=-5x+8
y=-5x/2+4
The y intercept of the 2nd equation is when x=0, so -7y=10
y=-10/7
The equation of the line would be y=-5x/2-10/7
Or in standard form....
5x/2+y=-10/7
5x+27=-20/7

5 0

3 years ago

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the sum of three number is 202, the second is three more than the first number, the third number is three times the second numbe

sertanlavr [38]

Answer:

x = 38

y = 41

z = 123

Step-by-step explanation:

x + y + z = =202

y = x + 3

z = 3y = 3(x + 3) = 3x + 9

<u>x</u> + <u>y</u> + <u>z</u> = <u>x</u> + <u>x + 3</u> + <u>3x + 9</u> = 5x + 12 = 202

5x + 12 = 202

5x = 190

x = 38

y = x + 3 = 41

z = 3y = 123

4 0

2 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago