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2 years ago

Expand and simplify (6x-2)^2

Mathematics

2 answers:

IgorLugansk [536]2 years ago

4 0

Answer:

$\sf 36x^2 -24x +4$

solving step wise:

$\sf (6x-2)^2$

$\sf (6x-2)(6x-2)$

$\sf 36x^2 -12x -12x +4$

$\sf 36x^2 -24x +4$

$\sf (6x-2)^2$

$\sf (6x)^2 +2*6x*-2+(-2)^2$

$36x^2 -24x +4$

ANTONII [103]2 years ago

3 0

Answer:

36x²-24x+4

Step-by-step explanation:

6x-2² 6x-2 6x-2

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Find the value of x (click photo)

Zolol [24]

Answer:

x = 130 degrees

Step-by-step explanation:

7-sided polygon = 900 degrees

114 + 138 + 127 + 108 + 142 + 141 + x = 900

770 + x = 900

-770 -770

x = 130

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5 0

2 years ago

Read 2 more answers

A manager wants to know if the mean productivity of two workers is the same. For a random selection of 30 hours in the past mo

vekshin1

Answer:

A.The data should be treated as paired samples. Each pair consists of an hour in which the productivity of the two workers is compared.

Explanation:

If the mean productivity of two workers is the same.

For a random selection of 30 hours in the past month, the manager compares the number of items produced by each worker in that hour.

There are two samples and the productivity of the two men is paired for each hour.

7 0

2 years ago

The sum of four consecutive even integers is -28

mina [271]

I guess you want to find the 4 numbers

x + x + 2 + x + 4 + x + 6 = -28
4x + 12 = - 28
4x = - 40
x = -10

so the numbers are -10, -8, -6 ,-4.

8 0

2 years ago

Smith's age is 8 and his mother is 42.in how many years will the mother be three times as his son

frozen [14]

Let’s make an equation
Let’s say x = amount of years after
Smith = 8, mother = 42
3(8 + x) = 42 + x
24 + 3x = 42 + x
2x = 18, x = 9
Solution: after 9 years

3 0

2 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

2 years ago

Expand and simplify (6x-2)^2​

Expand and simplify (6x-2)^2