3 years ago

20n^4+n^3+10n^2) / 10n^2

Mathematics

1 answer:

Zina [86]3 years ago

3 0

Divide them
We'd get
2n^2+1/10n + 1

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A rectangle has a perimeter of 84 meters and a length of 35 meters. What is the width of the rectangle?

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To find the perimeter of a rectangle, we can use this formula: $2L + 2W$
Where L is the length and W is the width.

We know the perimeter and we know the length and we want to find the width.
Substitute the values back into the formula.

2(35) + 2W = 84
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One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
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the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

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Answer:

Step-by-step explanation:

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