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3 years ago

12

Please help, I'm at my wits end :(.

1 answer:

Drupady [299]3 years ago

4 0

Answer:

im sorry theres not enough information for me to complete your request

Step-by-step explanation:

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sammy [17]

Answer:

3

Step-by-step explanation:

I can see it in the demonstration graph

8 0

3 years ago

Read 2 more answers

Day 4 of trying to find kiyah mccain :(

ratelena [41]

Answer:

who is that? Maybe i know this person...

Step-by-step explanation:

8 0

2 years ago

What decimal is halfway between 18 hundredths and 3 tenths

meriva

Answer:

its not a decimal but 915 is halfway between 18 hundredths and 3 tenths

Step-by-step explanation:

5 0

3 years ago

Pleeease open the image and hellllp me

Verdich [7]

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^{\ln x^x}=e^{x\ln x}$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of <em>y</em> is denoted <em>y' </em>)

$y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\dfrac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\dfrac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=\frac{\cos x}{\sin x}$ , we can cancel one factor of sine:

$y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^{\sin x}}$

$y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'$

$y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'$

$y'=e^{e^{\sin x}+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to <em>e</em>, you can use the change-of-base formula first:

$\log_2x=\dfrac{\ln x}{\ln2}$

Then

$(\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}$

So we have

$y=\cos^2(\log_2x)$

$y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'$

$y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'$

$y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

and we can use the double angle identity and logarithm properties to condense this result:

$y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

$\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'$

$2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0$

$-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)$

$y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}$

$y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}$

6. Same as with (5):

$\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'$

$\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)$

$y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}$

7. Looks like

$y=x^2-e^{2x}$

Compute the second derivative:

$y'=2x-2e^{2x}$

$y''=2-4e^{2x}$

Set this equal to 0 and solve for <em>x</em> :

$2-4e^{2x}=0$

$4e^{2x}=2$

$e^{2x}=\dfrac12$

$2x=\ln\dfrac12=-\ln2$

$x=-\dfrac{\ln2}2$

7 0

3 years ago

The expression ( x − 12 ) is a factor of x 2 + p x − 48 . What is the value of p ?

Anon25 [30]

Answer:

p = -8

Step-by-step explanation:

Since, (x - 12) is a factor, then;

x - 12 = 0

x = 12

substitute the value of x in the equation

$x^{2}$ + px - 48 = 0

$(12)^{2}$ + 12p - 48 = 0

144 + 12p -48

12p = -144 + 48

12p = -96

p = -8

So that,

$x^{2}$ +-8x - 48 = 0

$x^{2}$ -12x + 4X- 48 = 0

( $x^{2}$ -12x) + (4X- 48)

X(X - 12) + 4(X - 12)

(X - 12) (X + 4)

Therefore, p = -8 and (X - 12) (X + 4) are factors of the equation.

5 0

2 years ago