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2 years ago

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What is the simplified form of StartRoot StartFraction 72 x Superscript 16 Baseline Over 50 x Superscript 36 Baseline EndFractio

n EndRoot? Assume x ≠ 0.
StartFraction 6 Over 5 x Superscript 10 Baseline EndFraction
StartFraction 6 Over 5 x squared EndFraction
Six-fifths x Superscript 10
Six-fifths x squared

1 answer:

Iteru [2.4K]2 years ago

6 0

The simplification form of the provided expression is 6/5x¹⁰ option first is correct.

<h3>What is an expression?</h3>

It is defined as the combination of constants and variables with mathematical operators.

We have an expression:

$\rm = \sqrt{\dfrac{72x^{16}}{50x^{36}}}$

$\rm = \sqrt{\dfrac{72}{50}} \sqrt{\dfrac{x^{36}}{x^{16}}}$

$\rm = \sqrt{\dfrac{36\times2}{25\times2}} \sqrt{\dfrac{x^{36}}{x^{16}}}$

$\rm ={\dfrac{6x^{8}}{5x^{18}}}$

$\rm ={\dfrac{6}{5x^{10}}}$

Thus, the simplification form of the provided expression is 6/5x¹⁰ option first is correct.

Learn more about the expression here:

brainly.com/question/14083225

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Pls help asap<br> If g(x) is a linear function such that g(-3) = 2 and g(1) = -4, find g(7)

PilotLPTM [1.2K]

Answer:

G(x) = F(x) = 1/2x

Step-by-step explanation:

8 0

3 years ago

4|m−n| if m=−7 and n=2 i really need to know what this is asap please

vekshin1

Answer:

36

Step-by-step explanation:

Plug in -7 as m and 2 as n into the expression:

4 | m - n |

4 | -7 -2 |

Solve:

4 | -9 |

4(9)

= 36

6 0

3 years ago

(giving brainlelist if answered correctly!) which of the following inequalities is -4<br> ?

kicyunya [14]

-5 is always smaller than -4. So, x might be -4. So, A is an option you should select.
2 is greater than -4. So, x might be -4. So, B is an option you should select.
If we place -4 in x, we get 3(-4)+4, i.e., -12 + 4 = -8. So, 3x + 4 = -8. Therefore, you cannot select C because it cannot be greater than, it should be always equal to.
5(-4) = -20, 2(-4) - 3 = -8 -3 = -11. -11 is greater than -20 but they are never equal. So, you cannot select D.
4 - 2(-4) = 4 + 8 = 12. 12 is greater than -6. Therefore, you should select E.

<u>Answer</u>

<u>A,</u><u> </u><u>B,</u><u> </u><u>E.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

3 years ago

What did measure ST equal? please show your work so i can learn how to do this by myself <3

attashe74 [19]

253 I wish I could help with this problem but if there is similar problems I can help

5 0

3 years ago

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

3 years ago