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2 years ago

Please help me simplify the expression and please show work!!! <3

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

8 0

Step-by-step explanation:

$\frac{x + 4}{3 {x}^{2} - 12x - 96} = \frac{x + 4}{3( {x}^{2} - 4x - 32) } = \frac{x + 4}{3(x - 8)(x + 4)}$

$= \frac{1}{3(x - 8)} = \frac{1}{3x - 24}$

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How many £7.50 is there in £217.5?<br>

Soloha48 [4]

Answer:

Step-by-step explanation:

217.5÷7.50=29

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3 years ago

If all the dimensions of the prism doubled, would the volume double?

valentina_108 [34]

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3 years ago

Zachary is solving the following equation.

Monica [59]

Answer: C. Distribute 3/4 over (x-8)

Step-by-step explanation:

To solve any(technically most, but still...) algebraic equation you must isolate the variable. In this case, the first step would either be to divide both sides of the equation by 3/4, or distribute 3/4 to x and -8, as you cannot modify the contents of the parenthesis without removing the 3/4.

Hope it helps and lmk if you further explanation <3 :)

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3 years ago

Read 2 more answers

Consider the force field and circle defined below.

grin007 [14]

By Green's theorem,

$\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy$

where $C$ is the circle $x^2+y^2=9$ and $D$ is the interior of $C$ , or the disk $x^2+y^2\le1$ .

Convert to polar coordinates, taking

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

Then the work done by $\vec F$ on the particle is

$\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0$

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3 years ago

Please help, I really need this

Nady [450]

9514 1404 393

Answer:

(a) -- the correct choice is highlighted

Step-by-step explanation:

The units of specific heat tell you what quantities make up the ratio.

$\dfrac{390\text{ J}}{1\text{ kg$\cdot^\circ$C}}=\dfrac{-12.0\text{ J}}{0.012\text{ kg}\cdot\Delta T}\\\\\Delta T=\dfrac{-12.0}{0.012\cdot390}\ ^\circ\text{C}\approx-2.56\text{ $^\circ$C}$

The temperature will decrease by 2.56 C.

4 0

3 years ago