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3 years ago

I could use some help on this. thanks

Mathematics

2 answers:

AleksAgata [21]3 years ago

5 0

Answer: is B

Step-by-step explanation:

Dennis_Churaev [7]3 years ago

5 0

Answer:

1 to 1?

Step-by-step explanation:

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Find the circumference of the circle in terms of pi

madam [21]

It's 78π inches so probably d.

3 0

3 years ago

Helpppp meee pleaseee

miss Akunina [59]

B= the square root of 65

Formula: a^2= c*2-b^2

a^2= 81-16= 65

Once you find the square root of 65, thats your answer, but here they want you to keep it as 65.

Hope this helps!

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

Anyone understand how to do this??

Oxana [17]

Answer:

T=-11

Step-by-step explanation:

y2-y1/x2-x1

=t-(-1)/9-10

=t+1/9-10

=t+1/-1

=-11+1/1

=-10/1

=-10

Hope this helps :)

7 0

3 years ago

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Kendra is buying an outfit that is on sale for 20% off. The original price of the outfit is $109. What is the sale price? $130.8

aliya0001 [1]

$109 X 20%
$109 X 0.2=$21.8
$109-$21.8=$87.20

7 0

3 years ago

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