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PilotLPTM [1.2K]
3 years ago
5

Select the correct value.

Mathematics
2 answers:
Dafna1 [17]3 years ago
7 0

<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>2</em><em>0</em><em>.</em>

<em>look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em>

<em>hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>

dybincka [34]3 years ago
5 0

Answer:

20

Step-by-step explanation:

Given:  0.40x = 32, where x is the unknown number.

Then x = 32/0.40 = 80,

and 25% of that is

0.25(80) = 20

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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
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