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2 years ago

Do anyone know -8(1+4x)+7x=-25-8x

Mathematics

2 answers:

bulgar [2K]2 years ago

8 0

Answer:

x = 1

Step-by-step explanation:

-8(1+4x)+7x=-25-8x (Given)

-8 + -32x + 7x = -25 - 8x (Distribute)\

-8 + -25x = -25 - 8x (Combine like terms)

-17x = -17 (Move variables(x) to the left and constants to the right)

x = 1 (Divide) (the answer is positive because when you divide a negative number by a negative number it turns positive)

Arturiano [62]2 years ago

5 0

X=1 Same explanation of the first Who answered the problem

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Read 2 more answers

Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

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3 years ago

Suppose that many large samples are taken from a population and that the

rewona [7]

Answer:

$0.22 -1 *0.029 =0.191$

$0.22 +1 *0.029 =0.249$

And the best option would be:

D. (0.191 to 0.249)

Step-by-step explanation:

For this case we know that the mean is:

$\bar X = 0.22$

And the standard error is given by:

$SE = 0.029$

We want to construct a 68% confidence interval so then the significance level would be :

$\alpha=1-0.68 = 0.32$ and $\alpha/2 =0.16$ . The confidence interval is given by:

$\bar X \pm z_{\alpha/2} SE$

Now we can find the critical value using the normal standard distribution and we got looking for a quantile who accumulate 0.16 of the area on each tail and we got:

$z_{\alpha/2}= 1$

And replacing we got:

$0.22 -1 *0.029 =0.191$

$0.22 +1 *0.029 =0.249$

And the best option would be:

D. (0.191 to 0.249)

5 0

2 years ago