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Ronch [10]
3 years ago
12

Solve 9x is greater than or equal to 3

Mathematics
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

3 0
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.

5 0
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
4 0
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
3 0
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
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