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Oxana [17]
4 years ago
15

Suppose that a is a nonzero constant for which the equation ax^2+20x+7=0 has only one solution. find this solution.

Mathematics
1 answer:
Talja [164]4 years ago
4 0
\Delta=0\\\\
\Delta=20^2-4\cdot a\cdot7=400-28a\\
400-28a=0\\
28a=400\\
a=\dfrac{400}{28}=\dfrac{100}{7}


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Answer:

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Step-by-step explanation:

<u>This is a right triangle as shown by the right angle.</u>

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So we have the opposite of A and the hypotenuse.

The trigonometric ratio that relates opposite side to hypotenuse is SINE.

sin\theta=\frac{opposite}{hypotenuse}

Since, we need angle A, \theta is A, opposite side is 3, and hypotenuse is 9, we substitute:

sinA=\frac{3}{9}\\sinA=\frac{1}{3}

<em>To solve for the angle, we use our calculator and find the inverse sin of \frac{1}{3} :</em>

A=sin^{-1}(\frac{1}{3})\\A=19.47

<em>Rounding to nearest tenth of a degree:</em>

A=19.5

B is the right choice.

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2 years ago
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