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3 years ago

Select the graph of the function f (x) = 1/4 Then find the value of f (x) when x = -6.

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

7 0

The answer is c
And the answer is 9

gavmur [86]3 years ago

6 0

The answer is C good luck!!

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3 years ago

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Artie's brother is able to do 12 more push-ups than Artie. Artie represented the total number of push-ups his brother can do usi

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Step-by-step explanation:

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3 years ago

Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

$x = \frac{2}{3}$

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

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3 years ago

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if you earn 684.80 for a 32 hour work week, how much would you earn for a 40 hour workweek at the same hourly rate

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3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$