Use the quadratic formula to solve Round irrational solutions to the nearest tenth

Mathematics

1 answer:

Paha777 [63]4 years ago

7 0

Answer:

C. x = -5.6 and x = -1.4

Step-by-step explanation:

The quadratic formula is $x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ . Here a=1, b=7, and c=8.

Substitute and you'll have:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a} =\frac{-7+/-\sqrt{7^2-4(1)(8)} }{2(1)}=\frac{-7+/-\sqrt{49-32} }{2)}$

$\frac{-7+/-\sqrt{17} }{2}=\frac{-7+/-4.1 }{2}= -1.4, -5.6$

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Svetradugi [14.3K]

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The side opposite of 90* should be the largest. The size opposite of 55 would be the second largest one. The side opposite of 35 would be the smallest one. Using pythagoras theorem again, I can try plugging in both 11.47 and 16.38 with c^2 - b^2 = a^2

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400 - 132 =268 (sq root now = 16.37) this would mean that A is larger than B. B should be larger than A. Try the other one

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400 -268 = 132 (sq root = 16.37) this would mean that B is larger than A which is what we want.

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7 0

3 years ago

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klemol [59]

Answer: Third option.

Step-by-step explanation:

By definition, Exponential functions have the following form:

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It is importat to remember that the "Zero exponent rule" states that any base with an exponent of 0 is equal to 1.

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