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3 years ago

Explain the steps when solving this. 10 + h > 2 + 2h

2 answers:

5 0

Solve it like you would for an equation by isolating the variable h.

First subtract h from both sides.

10+h-h>2+2h-h
10>2+h

Then subtract 2 from both sides.

10-2>2-2+h
8>h

We can then flip 8>h to h<8 just because it's a more standard way of writing it.

Final answer: h<8

kherson [118]3 years ago

3 0

you first subtract from both sides and get 10 > 2 + h then from both sides subtract 2 to get 8>h and written a standard way, the correct answer would be h<8 (so basically just flip 8>h)

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2 years ago

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3 years ago

Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
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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3 years ago