If a and b are real numbers such that a > 0 and b < 0, then which of the following is equivalent

Mathematics

1 answer:

Pepsi [2]3 years ago

7 0

27 I think I hope this help u

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Where does the section of the bridge meet ground level? Solve 0 = -x2 + 10x - 8

djverab [1.8K]

In can be less mental effort if the equation is multiplied by -1 first.
.. x^2 -10x +8 = 0
.. x^2 -10x +25 +8 -25 = 0
.. (x -5)^2 = 17
.. x -5 = ±√17
.. x = 5 ±√17

The values of x that satisfy the given equation are
.. x = 5 -√17 ≈ 0.877
.. x = 5 +√17 ≈ 9.123

4 0

3 years ago

Read 2 more answers

PROVE THAT:<br><br>cos 20° - sin 20° = \sqrt{2}sin25°<br><br>

Lemur [1.5K]

Answer:

See below.

Step-by-step explanation:

$\cos(20)-\sin(20)=\sqrt{2}\sin(25)$

First, use the co-function identity:

$\sin(90-x)=\cos(x)$

We can turn the second term into cosine:

$\sin(20)=\sin(90-70)=\cos(70)$

Substitute:

$\cos(20)-\cos(70)=\sqrt{2}\sin(25)$

Now, use the sum to product formulas. We will use the following:

$\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$

Substitute:

$\cos(20)-\cos(70)=-2\sin(\frac{20+70}{2})\sin(\frac{20-70}{2})\\\cos(20)-\cos(70) =-2\sin(45)\sin(-25)\\\cos(20)-\cos(70)=-2(\frac{\sqrt{2}}{2})\sin(-25)\\ \cos(20)-\cos(70)=-\sqrt{2}\sin(-25)$