If a secant<span> and a </span><span>tangent of a circle </span><span>are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.
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y</span>² = 7(15+7)
<span>y</span>² = 7*22
<span>y</span>² = 154
<span>y = </span>√154
<span>y = 12.4 </span>← to the nearest tenth<span>
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Answer:
Step-by-step explanation:
Hello,
Pls post the full question next time so that we can make sure what is expected.
I assume that you want to solve for (x,y)
(1) 2x + 3y = 7
(2) 3x + 2y = 8
2*(2)-3*(1) gives
2*3x + 2*2y - 3*2x - 3*3y = 2*8 - 3*7 = 16 - 21 = -5
6x - 6x + 4y -9y = -5
-5y=-5
y = 1
and we can replace in (1)
2x + 3*1 = 7
2x = 7 - 3 = 4
x = 2
hope this helps
Step-by-step explanation:
Area = 8 × 5
Area = 40^2
Perimeter is the length around the edges
Therefore it is 8+8+5×5 (since there are 4 edges)
Perimeter = 2×8 + 2×5
= 40 + 10
= 50cm
180 is 6x larger than 30, and 30 is 6x larger than 5. so y is 6x larger than x, and n is 30
10. If it rained 7/10 of the days, 3/10 days did not receive rain.
90 days multiplied by 3/10 is 27.
It would not rain about 27 days in the 90 days.
11. x multiplied by 8 and a half equals 5.
(8 and 1/2)x = 5
x = .588235 which is about 6/10
12. C = 6/10 and D = 8/10 so the product would be 48/100 which is close to 5/10. This point is N.
