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3 years ago

(x^3−7x−6) : (x−4) Kan someone solve this for me, and show how to get it?

Mathematics

1 answer:

Vladimir79 [104]3 years ago

3 0

Answer:

x = 3 or x = -1 or x = -2

Step-by-step explanation:

Solve for x:

(x^3 - 7 x - 6)/(x - 4) = 0

Hint: | Multiply both sides by a polynomial to clear fractions.

Multiply both sides by x - 4:

x^3 - 7 x - 6 = 0

Hint: | Factor the left hand side.

The left hand side factors into a product with three terms:

(x - 3) (x + 1) (x + 2) = 0

Hint: | Find the roots of each term in the product separately.

Split into three equations:

x - 3 = 0 or x + 1 = 0 or x + 2 = 0

Hint: | Look at the first equation: Solve for x.

Add 3 to both sides:

x = 3 or x + 1 = 0 or x + 2 = 0

Hint: | Look at the second equation: Solve for x.

Subtract 1 from both sides:

x = 3 or x = -1 or x + 2 = 0

Hint: | Look at the third equation: Solve for x.

Subtract 2 from both sides:

Answer: x = 3 or x = -1 or x = -2

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Brie knows there is a relationship between the grades on her math tests and the time she spends studying for those test. Identif

erma4kov [3.2K]

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Step-by-step explanation:

The relationship between Dependent and independent variables lies in the fact that, the outcome (dependent variable) relies on the value of the independent variable. Hence, the independent variable could be explained as a series of different possible values which has an effect on the value of the predicted variable. In an experiment, the dependent variable is predicted based on the input or value of the independent variable entered into the model.

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3 years ago

Read 2 more answers

What is an extraneous solution to the equation b/b-3 -5/b=3/b-3 ?

storchak [24]

$\frac{b}{b-3} - \frac{5}{b} = \frac{3}{b-3} \\ \frac{ b^{2} -5(b-3)}{b(b-3)} =\frac{b}{b-3} \\ \frac{ b^{2} -5b+15}{b^2-3b} =\frac{b}{b-3} \\ (b-3)(b^{2} -5b+15)=b(b^2-3b) \\ b^3-8b^2+30b-45=b^3-3b^2 \\ 5b^2-30b+45=0 \\ b^2-6b+9=0 \\ (b -3)(b-3)=0 \\ b=3$

To check: 3/3-3 - 5/3 = 3/0 - 5/3 which is not defined.

Therefore, b = 3 is an extraneous solution.

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3 years ago

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Nutka1998 [239]

Answer:

2.75

step-by-step explanation:

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3 years ago

WILL GIVE BRAINLIEST: Which function has a removable discontinuity?

elena55 [62]

Answer:

First choice

Step-by-step explanation:

The discontinuity is removable if by reducing the fraction that discontinuity doesn't continue to exist as a discontinuity.

Example of, (x-1)/(x-1) has a a discontinuity at x=1 and it's removable because the fraction reduces to 1 which doesn't have a discontinuity at x=1.

Example not of, (x-1)/(x-2) has a discontinuity at x=2 and it is not removable because we can't get rid of the x-2 factor in the denominator.

The first choice has a discontinuity at x=-1 and it is removable because x^2-x-2=(x-2)(x+1) and the x+1's will cancel on top and bottom making the point at x=-1 a removable discontinuity.

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2 years ago