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Vera_Pavlovna [14]

3 years ago

5

H() =

https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex" id="TexFormula2" title="(\frac{1}{2})^x" alt="(\frac{1}{2})^x" align="absmiddle" class="latex-formula">
Find the value of $x$ when h( $x$ ) = 8.

1 answer:

Yuliya22 [10]3 years ago

6 0

Answer:

x = - 3

Step-by-step explanation:

Solving

$(\frac{1}{2}) ^{x}$ = 8

$(2^-1)^{x}$ = 2³

$2^{-x}$ = 2³

Since the bases on both sides are equal equate the exponents

- x = 3 ( multiply both sides by - 1 )

x = - 3

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3 years ago

4⁄8 + ¼ = ? What would the answer be?

Lorico [155]

3/4. 4/8= 2/4. so 1/4+2/4=3/4

3 0

3 years ago

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If carlos did some work and got 3x + 2x - 6 = 24 what mistake did he make

Makovka662 [10]

Carlos made the mistake that he did not combine like terms (3 x and 2 x) properly and did not use addition property of equality.

<u>Step-by-step explanation:</u>

Carlos did the work as 3 x + 2 x - 6 = 24

We need to find his mistake that he made in above given.

Here, he did not add the like terms (3 x and 2 x)

3 x + 2 x = 5 x

Therefore, his work should be

5 x - 6 = 24

Also, he did not use addition property of equality. It means the equation remains same even though the same number gets added on both sides. It would be

5 x - 6 = 24

+ 6 = + 6

-----------------------

5 x = 30

Dividing 30 by 5, we get answer as '6'. Hence,

$x=\frac{30}{5}$ = 6

So, stated the above two are the mistakes found in carlos work.

4 0

3 years ago

(x^2+x-17)/(x-4) solve this

sergij07 [2.7K]

Answer:

$\displaystyle x + 5 + \frac{3}{x - 4}$

Step-by-step explanation:

Since the divisor of $\displaystyle x - 4$ is in the form of $\displaystyle x - c,$ we use what is called Synthetic Division. Now, in this formula, −c gives the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

4| 1 1 −17

↓ 4 20

_________

1 5 3 → $\displaystyle x + 5 + \frac{3}{x - 4}$

You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x² + x - 17]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x, the 5 follows right behind it, and bringing up the rear, $\displaystyle \frac{3}{x - 4},$ giving you the quotient of $\displaystyle x + 5 + \frac{3}{x - 4}.$ However, in this case, since you have a remainder of 3, this gets set over the divisor.

I am joyous to assist you anytime.

8 0

3 years ago

For what value of a should you solve the system of elimination?

SIZIF [17.4K]

$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

Simplify more.

$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

$20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}$

$\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10$

$Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20$

$\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more$

$10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite$
$-3a+2\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine$
$60\left(-a+2\right)$

$\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \ \frac{10\left(-a+2\right)}{\left(-3a+10\right)}$

$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers