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ladessa [460]
4 years ago
10

7/2x-8=3/x-6 What does x equal?

Mathematics
1 answer:
Burka [1]4 years ago
5 0

Answer:

x=2+ sqrt46/7

x=2-sqrt46/7

Step-by-step explanation:

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45 points to answer this question
Mrac [35]

if a = b and b = c, then a = c.

In this case if y = - 7 and -7 = z then y = z  .................>(y = a, -7 = b and z = c)

Answer:  B)

transitive property of equality




7 0
3 years ago
The shoe size for all the pairs of shoes in a person's closet are recorded.
Dovator [93]
The mean/average is found by adding all of your values and then dividing by the number of values found.

7+7+7+7+7+7+7+7+7+7=70

There are 10 size 7 shoes. Divide 70 by 10 to get the mean, which is 7.

Another ex.

Find the average of numbers listed

1+4+8
these add up to 13
there are 3 numbers listed
13/3=4.33 so the mean is 4.33
5 0
3 years ago
How many sevens are there is 3500
Marrrta [24]

Answer:

there are no seven in there

7 0
3 years ago
Read 2 more answers
Please help me! Thank you. :)
amm1812

Answer:

13.75/ hour

Step-by-step explanation:

$192.50/14 hours= $13.75/hour

$247.50/18 hours= $13.75/hour

$453.75/33 hours= $13.75/hour

5 0
3 years ago
Read 2 more answers
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3
(x^{1/2}-x)(x^{1/2}+x)=x-x^2

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
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