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Studentka2010 [4]

3 years ago

8

Graph the solution to the following system of inequalities in the coordinate plane.

1 answer:

Goshia [24]3 years ago

4 0

Answer:

see below

Step-by-step explanation:

The first equation is in slope-intercept form, so you can see that the boundary line has a slope of -2 and goes through the point (x, y) = (0, -4). Since the comparison is "<", the line is dashed and shading is below it.

The second equation is that of a vertical boundary line at x=-3. It is solid, because the comparison includes the "equal" case. Shading is to the right of it, where x values are greater than -3.

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#3 please im so bad with area lol

ANTONII [103]

Base times height is the formula for area I’m pretty sure

4 0

3 years ago

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I need help asap someone please help me i dont understand this question so can someone help me

eduard

Answer:

we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

Step-by-step explanation:

Given the expression

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

$=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}$

$=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}$

cancel the common factor: 2

$=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}$

cancel the common factor: p

$=\frac{2p+1}{p^2\left(4p^2-1\right)}$

$=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}$

cancel the common factor: 2p+1

$=\frac{1}{p^2\left(2p-1\right)}$

Expanding

$=\frac{1}{2p^3-p^2}$

Thus, we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

7 0

3 years ago

Need help solving for X

Bond [772]

X= -2

look at the 6x-12 you divide -12 by 6 and get x=-2

6 0

3 years ago

The radius of a circle is 13miles. what is the area of a sector bounded by a 20 degree arc

Stolb23 [73]

Answer:

Area of the sector = 29.48 square miles.

Step-by-step explanation:

Given:

Radius of a circle = 13 miles

Angle bounded by the arc = 20 deg

To find

Area of the sector bounded by 20 degrees.

Note: Area of a sector = $\frac{(\theta)}{360}\times \pi\times r^2$

Plugging the values in the equation.

⇒ $\frac{(\theta)}{360}\times \pi\times r^2$

⇒ $(\frac{20}{360})\times 3.14\times 13^2$

⇒ $29.48$ square-miles

So the area of the sector bounded by 20 degree arc is 29.48 square miles.

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3 years ago

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Musya8 [376]

Answer 1061

Step-by-step explanation:

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