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3 years ago

Solve for x.show your work please - 1/2x < -12

Mathematics

2 answers:

Nastasia [14]3 years ago

6 0

Greetings.

The answer is x > 24.

Explanation:

Solving Inequality is almost the same as equation. However, you have to know one thing about solving an Inequality.

That is whenever you move the negative coefficient of the variable, you have to swap the sign of Inequality.

$-\frac{1}{2}x$

Reminder that

$-\frac{1}{2}x=-\frac{x}{2}$

We use negative x above 2 instead (Easier in solving)

$-\frac{x}{2}$

Then we move 2 to divide -12.

$-x$

Because the coefficient for x is in negative. We move -1 to divide -24 as we get 24 (Because negative divide negative = positive)

But remember that "whenever you move the negative coefficient of the variable, you have to swap the sign of Inequality."

So we change from < to > instead.

$x>24$

Ratling [72]3 years ago

3 0

Answer:

x <24

Step-by-step explanation:

first cancel negative signs

then transfer 2 to right side

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3 years ago

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Andreas93 [3]

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3 years ago

Read 2 more answers

What is the equation of the function that is graphed as line a?

valentinak56 [21]

F(x)=-1x-1

1) Pick 2 points on the line (i chose (-4,3) and (3,-4)
2) Find slope of the line using the 2 points. (work below)
3) Find y-intercept, which is the point where y-axis and the line cross (y-intercept is -1).
4) Place both slope and y-intercept into slope-intercept form, y=mx+b (m=slope and b=y-intercept.)
5) Change y to f(x) (meaning function of x).

Work:
3-(-4)/-4-3=3+4/-4-3=7/-7=-1

y-intercept equals -1 also.

y=mx+b
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4 years ago

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

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