<img src="https://tex.z-dn.net/?f=6%20%3D%20%20%20%5Cfrac%7Ba%7D%7B4%7D%20%20%2B%202" id="TexFormula1" title="6 = \frac{a}{4}

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Artyom0805 [142]

4 years ago

+ 2" alt="6 = \frac{a}{4} + 2" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

kvv77 [185]4 years ago

5 0

Multiply by 4 on both sides. This results into:

24 = a + 8

If 8 is substracted in both sides then a would be:

16 = a

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All of the following are equivalent to 2(2a + b) + 8, except _____.

Nady [450]

Where are the choices to the problem??

7 0

3 years ago

Read 2 more answers

Find the angle between vector u=2i+sqrt(11)j and v=-3i-2j to the nearest degree

const2013 [10]

Answer: Last option 155°

Step-by-step explanation:

We have the components of the vectors u and v.

Then, to find the angle between them, perform the following steps:

1) Calculate the scalar product u * v

If $u = 2i + \sqrt{11}j$ and $v = -3i-2j$

Then, the product scalar u * v is:

$u * v = (2)(-3) + (\sqrt{11})(-2)$

$u * v = -6-2\sqrt{11}$

$u * v = -12.633$

2) Calculation of the magnitude of both vectors.

$| u | = \sqrt{(2) ^ 2 + (\sqrt{11})^2}\\\\| u | = \sqrt{4+11}\\\\| u | = \sqrt{15}$

$| v | = \sqrt{(- 3) ^ 2 + (- 2) ^ 2}\\\\| v | = \sqrt{9 +4}\\\\| v | = \sqrt{13}$

3) Now that you know the product point between the two vectors and the magnitude of each, then use the following formula to find an angle

$u * v = | u || v |cos(\alpha)$

$-12.633 = \sqrt{15}\sqrt{13}*cos(\alpha)\\\\cos(\alpha) = \frac{-12.633}{\sqrt{15}\sqrt{13}}\\\\arcos(\frac{-12.633}{\sqrt{15}\sqrt{13}}) = \alpha\\\\\alpha =155\°$

7 0

4 years ago

Can you pleaseee help meee

Mrac [35]

Yesssssssss, with what though

6 0

3 years ago

Lets find 1/3 - 1/12<br><br> 1/3 - 1/12 = ___/12 - 1/12 =___/___<br><br> Fill in the blanks

kicyunya [14]

Jacks mama jsnsnakkanama sokzna

4 0

3 years ago

Write an expression that is equivalent to (-16)^-2

raketka [301]

Answer:

1/256

Step-by-step explanation:

(-16)^-2

= 1/(-16)^2

= 1/ (-16· -16)

= 1/ 256

8 0

3 years ago