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Brilliant_brown [7]
3 years ago
11

What are the solutions to the equation (x-6)(x+8)=0?

Mathematics
1 answer:
Zepler [3.9K]3 years ago
6 0

Answer:

x = 6, -8

Step-by-step explanation:

If (x - 6)(x + 8) = 0, that would imply that either (x - 6) or (x + 8) would equal zero. Using this, we can find that solving the two equations:

x - 6 = 0

and

x + 8 = 0

would yield the two solutions to the equation.

x - 6 = 0

Add 6 to both sides of the equation.

x = 6

So one of the solutions would be x = 6.

x + 8 = 0

Subtract 8 from both sides of the equation.

x = -8

So the other solution would be x = -8.

The two solutions are x = 6 and x = -8.

I hope you find my answer and explanation to be helpful. Happy studying.

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Which expressions is equivalent to 2r+(t+r)
azamat

Answer:

t+3r

Step-by-step explanation:

This is a one step solution

All you do is add like terms

In this case the only like terms are the varibles r

2r+(t+r)

so if you add all of the rs together there would be 3rs

So the rest of the problem would continue as normal while bringing the 3r to the end

t+3r

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3 years ago
How would you know the answer (important)
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Answer:

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The answer
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3 years ago
Is √3/16 rational or irrational
kotykmax [81]

Irrational number because it cannot be expressed as a fraction of two whole numbers and it has no accurate decimal equivalent.

So its  Irrational hope this helped :)

6 0
3 years ago
If we approximate the function y=sin(x) with a0+a1 x a2 x^2 +a3 x^3, what is a0,a2,a2,a3?
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The coefficients a_0,a_1,a_2,a_3 could be chosen to be the coefficients in the Maclaurin series of \sin(x).

We have

y = \sin(x) \approx a_0 + a_1 x + a_2 x^2 + a_3 x^3 \\\\ \implies y(0) = 0 = a_0

y' = \cos(x) \approx a_1 + 2a_2 x + 3a_3 x^2 \\\\ \implies y'(0) = 1 = a_1

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