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4 years ago

Solve for x. 6(x - 2) = 4 x = 1 x = 1 1/3 x = 2 2/3

Mathematics

2 answers:

ElenaW [278]4 years ago

8 0

It would be x = 2 2/3 because

2 2/3 - 2 = 2/3

6 x 2/3 = 4

Snezhnost [94]4 years ago

5 0

This is the correct answer!!! Hope this help

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Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Igoryamba

Hihi!

2/3 would be the right answer! Because 6 is 2/3 of 9.

I hope I helped!
-Jailbaitasmr

3 0

4 years ago

How to solve (3m+7)(2m-1)

Bad White [126]

the answer is

6m^2+11m-7

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5 0

3 years ago

What completes this expression to make it a perfect square trinomial? x2 + 4x + ___ A 2 B 4 C 8 D 16

brilliants [131]

Answer:

Option B. 4

Step-by-step explanation:

The given expression is x² + 4x + -----

We have to find a number which can be added in the expression to make it a perfect square trinomial.

x² + 2(2x) + 2²

= (x + 2)²

Therefore, if we add 4 in the given expression it becomes a perfect trinomial.

Option B is the correct option.

3 0

4 years ago

To factor 9x2 - 4, you can first rewrite the expression as:

MArishka [77]

Answer:

Option C is correct.

(3x)² - 2²

4 0

3 years ago

Read 2 more answers

Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π

Natasha2012 [34]

Let $H$ denote the helicoid parameterized by

$\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k$

for $0\le u\le1$ and $0\le v\le9\pi$ . The surface area is given by the surface integral,

$\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

We have

$\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j$
$\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k$
$\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k$
$\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}$

So the area of $H$ is

$\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$
$=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2$

5 0

3 years ago