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3 years ago

12

Solve 9x is greater than or equal to 3

1 answer:

hjlf3 years ago

4 0

Answer:

9999999999 9X GREATER THAN OR EQUAL TO 33333333333333

Step-by-step explanation:

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-6x-y=11<br> What’s the answer?

Serhud [2]

Answer:

_6x-y=11/ 6x+y+11=0 this is the answer

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3 years ago

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Help me ASAP PLEASE

tamaranim1 [39]

Answer:

Your(my) number is 40

Step-by-step explanation:

I. Given 80/something, + 13, * 5 = 75 when x is something

So [ $\frac{80}{x}$ + 13 ]5 = 75

[ $\frac{400}{x}$ + 65] = 75

$\frac{400}{x}$ = 75 - 65

$\frac{400}{x}$ = 10

400 = 10x

x = 40

II. Prove by replace 40 in x

[(80/40) + 13]5 = 75

[2 + 13] 5 = 75

(15)5 = 75

75 = 75 true

Ask me for anything.

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2 years ago

About your photograph 4 in wide and 6in long the photograph is enlarged and their larger is proportional to the original photogr

Bad White [126]

18 inches.

12/4=3 so you would multiply 6•3 to get 18

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3 years ago

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If the width of a square piece of material is 3x – 2, what is the area?

Zinaida [17]

The answer is B because its a square so all sides are equal and then you factor

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3 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