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3 years ago

Which of the following is the correct factorization of 27x³ - 8?

Mathematics

2 answers:

anyanavicka [17]3 years ago

6 0

(3x-2)(9x²+6x+4) I think that's it

bekas [8.4K]3 years ago

3 0

Answer:The second choice

Step-by-step explanation:

You might be interested in

Help!! Which number is NOT in scientific notation?

ki77a [65]

D is not in correct scientific notation. To be correct, the answer would be 5.0 x 10^2

If this helped you, please list me as brainliest!

4 0

3 years ago

The average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $138.00. Assuming a normal distribution, if 77 percent

jek_recluse [69]

Answer:

22.97

Step-by-step explanation:

Given that the average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $138.00.

Let x be the average cost an IRS Form 1040 tax filing at Thetis Tax Service

$P(X$ is given

From std normal distribution table we find 77th percentile z value.

z=0.74

Corresponding X value = 155

i.e. $0.74=\frac{155-138}{s}$ where s is the std deviation

Simplify to get

Std deviation = $\frac{17}{0.74} \\=22.97$

8 0

3 years ago

4x<br>f(x)=<br>x -3 togetherness

frozen [14]

Answer:

f(x)=2(\dfrac{2}{3})^x.

Step-by-step explanation:

f(x)=1/2(2)xf(x)=3/4(-1/5)xf(x)=(7/2)xf(x)=2(2/3)x

3 0

3 years ago

If two supplementary angles are in the ratio of 4:5 the measure of the larger angle is

dimulka [17.4K]

Answer:

Step-by-step explanation:

Let the angles be 4x , 5x

4x + 5x = 180 { supplementary}

9x = 180

x = 180/9

x = 20

Larger angle= 5x = 5* 20 = 100

6 0

3 years ago

If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

$= \dfrac{-1}{x(x+h)}, h\ne 0$

5 0

3 years ago