Pleeease open the image and hellllp me

Answers

Answer 1

Answer:

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 y is denoted y' )

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

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

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

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

2. Chain rule:

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

$y'=\frac1{\csc(3x)}(\csc(3x))'$

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

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

Since $\cot x=(\cos x)/(\sin x)$ , we can cancel one factor of sine:

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

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 e, you can use the change-of-base formula first:

$\log_2x=(\ln x)/(\ln2)$

Then

$(\log_2x)'=\left((\ln x)/(\ln 2)\right)'=\frac1{\ln 2}$

So we have

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

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

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

$y'=-\frac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

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

$y'=-\frac1{\ln2}\sin(2\log_2x)=-\frac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

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

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

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

$y'=(2x+\cos x\,e^y\ln y)/(2y-\sin x\,e^y-\frac xy)$

$y'=(2xy+\cos x\,ye^y\ln y)/(2y^2-\sin x\,ye^y-x)$

6. Same as with (5):

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

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

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

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

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

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 x :

$2-4e^(2x)=0$

$4e^(2x)=2$

$e^(2x)=\frac12$

$2x=\ln\frac12=-\ln2$

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

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What is the probability of selecting a cherry piece? Write answer as a reduced fraction.

Answers

Answer:6/14 or to round it up by 1/14 and make it simplified it would be 1/2

Step-by-step explanation:

One normally distributed with mean value 20 in. and standard deviation .5 in. The length of the second piece is a normal rv with mean and standard deviation 15 in. and .4 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between 34.5 and 35 in.

Answers

Answer:

The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Step-by-step explanation:

Let the random variable X represent the length of the first piece, Y represent the length of the second piece and Z represents the overlap.

It is provided that:

$X\sim N(20,\ 0.50^(2))\nY\sim N(15,\ 0.40^(2))\nZ\sim N(1,\ 0.10^(2))$

It is provided that the lengths and amount of overlap are independent of each other.

Compute the mean and standard deviation of total length as follows:

$E(T)=E(X+Y-Z)\n=E(X)+E(Y)-E(Z)\n=20+15-1\n=34$

$SD(T)=√(V(X+Y-Z))\n=√(V(X)+V(Y)+V(Z))\n=\sqrt{0.50^(2)+0.40^(2)+0.10^(2)}\n=0.6480741\n\approx 0.65$

Since X, Y and Z all follow a Normal distribution, the random variable T, representing the total length will also follow a normal distribution.

$T\sim N(34, 0.65^(2))$

Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:

$P(34.5<T<35)=P((34.5-34)/(0.65)<(T-\mu_(T))/(\sigma_(T))<(35-34)/(0.65))\n\n=P(0.77<Z<1.54)\n\n=P(Z<1.54)-P(Z<0.77)\n\n=0.93822-0.77935\n\n=0.15887\n\n\approx 0.1589$

*Use a z-table.

Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Need answer to. Add 3x-3and4x^2-6x

Answers

3x-3+4x^2-6x
4x^2-6x+3x-3 (arrange them in order)
4x^2-3x-3

do u want more simplified form ?

On Wednesday, a local hamburger shop sold a combined total of 360 hamburgers and cheeseburgers. The number of cheeseburgers sold was two times the number of hamburgers sold. How many hamburgers were sold on Wednesday?

Answers

Answer:

120

Step-by-step explanation:

360 combined, 2/3 is cheese, so 1/3 is hamburger, 1/3 of 360 is 120.

Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 64x2 + 81y2 = 1. $ L=\iint_{R} {\color{red}9} \sin ({\color{red}384} x^{2} + {\color{red}486} y^{2})\,dA $.

Answers

$\displaystyle\iint_R\sin(384x^2+486y^2)\,\mathrm dA$

Notice that Given that $R$ is an ellipse, consider a conversion to polar coordinates:

$\begin{cases}x(r,\theta)=\frac r8\cos\theta\ny(r,\theta)=\frac r9\sin\theta\end{cases}$

The Jacobian for this transformation is

$J=\begin{bmatrix}\frac18\cos\theta&-\frac r8\sin\theta\n\frac19\sin\theta&\frac r9\cos t\end{bmatrix}$

with determinant $\det J=\frac r{72}$

Then the integral in polar coordinates is

$\displaystyle\frac1{72}\int_0^(\pi/2)\int_0^1\sin(6r^2\cos^2t+6r^2\sin^2t)r\,\mathrm dr\,\mathrm d\theta=\int_0^(\pi/2)\int_0^1r\sin(6r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{(\pi\sin^23)/(864)}$

where you can evaluate the remaining integral by substituting $s=6r^2$ and $\mathrm ds=12r\,\mathrm dr$ .

Final answer:

To evaluate the integral, we make a change of variables using the transformation x=u/8 and y=v/9 to transform the region into a unit circle. Then we convert the integral to polar coordinates and evaluate it.

Explanation:

To evaluate the given integral, we can make the appropriate change of variables by using the transformation x = u/8 and y = v/9. This will transform the region R into a unit circle. The determinant of the Jacobian of the transformation is 1/72, which we will use to change the differential area element from dA to du dv. Substituting the new variables and limits of integration, the integral becomes:

L = \iint_{R} 9 \sin (612 u^{2} + 768 v^{2}) \cdot (1/72) \,du \,dv

Next, we can convert the integral from Cartesian coordinates(u, v) to polar coordinates (r, \theta). The integral can be rewritten as:

L = \int_{0}^{2\pi} \int_{0}^{1} 9 \sin (612 r^{2} \cos^{2}(\theta) + 768 r^{2} \sin^{2}(\theta)) \cdot (1/72) \cdot r \,dr \,d\theta

We can then evaluate this integral to find the value of L.

Learn more about Evaluation of Integrals here:

brainly.com/question/32205191

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D(n)=8−6(n−1) What’s the 6th term

Answers

D(n)=8−6(n−1). The 6th term is found by subbing 6 for n in this formula:

D(6)=8−6(6−1) = 8 - 6(5) = 8 - 30 = -22 (answer)

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