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Ira Lisetskai [31]

3 years ago

7

At the beginning of summer, the water level on a pond is 2 feet below its normal level. After an unusually dry summer, the water

level on the pond dropped another 2 feet. How far above or below the normal water level is the pond at the end of the summer?

1 answer:

Scilla [17]3 years ago

8 0

Answer:

Step-by-step explanation:

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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

6 0

3 years ago

Solve by using the perfect squares method. x2 + 4x + 4 = 0<br> a. 2<br> b. –2<br> c. 4<br> d. –4

Assoli18 [71]

Solve your equation step-by-step.

x2+4x+4=0

Factor left side of equation.

(x+2)(x+2)=0

Set factors equal to 0.

x+2=0

or

x+2=0

x=−2

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3 years ago

Read 2 more answers

Feliz [49]

Answer:

PRT%

Step-by-step explanation:

<em>PR</em><em>T</em>

<em> $= 320 \times 2.3 \times 20 \div 100 = 1472$ </em>

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At a hardware store they can cut a rope off of a big roll so you can buy any length you like. the cost of 6 feet of rope is 7.50

konstantin123 [22]

Answer:

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The lateral area that Rachel needs to cover is 21 inches.

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