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3 years ago

Simplify the following expression: -6x + 4(-x - 3)

Mathematics

1 answer:

ss7ja [257]3 years ago

3 0

Answer:

$- 10x - 12$

Step-by-step explanation:

$- 6x + 4( - x - 3) \\ - 6x - 4x - 12 \\ - 10x - 12 \\ = - 2(5x + 6)$

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Tennessee Whisky contains 40% alcohol; how many ml of whiskey can be produced with 80ml of alcohol?

anygoal [31]

well we know that we have 80mL and we also know that those 80mL will represent the "solute" or the 40% of the Tennessee whiskey total which we can say is 100%, so if 80 is the 40%, how much will it be for the 100% in mL?

$\begin{array}{ccll}mL&\%\\\cline{1-2}80&40\\x&100\end{array}\implies \cfrac{80}{x}=\cfrac{40}{100}\implies \cfrac{80}{x}=\cfrac{2}{5}\implies 400=2x\\\\\\ \cfrac{400}{2}=x\implies \underset{mL}{200}=x$

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3 years ago

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tangare [24]

Answer: ?

Step-by-step explanation:

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3 years ago

david is 8 years older than stephen and 2 years younger than Graham. if Neil is 7 years younger than Graham, how much older than

ANEK [815]

5 years older...cause if David is 2 years younger than Graham & Neil is 7 years younger than Graham...7-2=5 years :)

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3 years ago

PLEASE HELP ASAP…. THIS TEST IS MOST OF MY GRADE

Lemur [1.5K]

Answer:

The answer is D

Step-by-step explanation:

This is so because the opposite of a positive number is that number but negative. Hope this helps!

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3 years ago

Read 2 more answers

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

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3 years ago