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3 years ago

If (4 to the 4 power) to the x power = 4 to the 32 power, what is the value of x?

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

(4 to the 4 power) to the x power = 4 to the 32 power

(4^4)^x = 4^32

4^4x = 4^32

4x = 32

x = 8

Answer

x = 8

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Which equation’s graph is a horizontal line? *<br> x = 9<br> 3x + 2y = 8<br> y = - 5

Natasha2012 [34]

Answer:

Your answer would by y = -5

Step-by-step explanation:

This is horizontal because it does not have a slope. It is a horizontal straight line at the -5 point on the y-axis

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3 years ago

Find an equation of the plane that is parallel to the xz-plane and is located 11 units to the right of the xz-plane in standard

mina [271]

Answer:

y = 11

Step-by-step explanation:

Since our plane is parallel to the xz plane, the x and z are independent of each other, i.e. they could be anything. So xz plane has an equation of y = 0. So if this plane is located 11 units to the right of the xz plane then its equation must be y = 11.

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4 years ago

The next model of a sports car will cost 5.6% less than the current model. The current model costs $52,000. How much will the pr

Lostsunrise [7]

Answer:

49,088

Step-by-step explanation:

First you need to find 5.6% of 52,000. Simply multiply 52,000 by 0.056 (change the percent to a decimal) then you get 2,912.

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8 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

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3 years ago

Y=-5x – 26<br> 15y=3x – 18

harina [27]

Answer:

Y = − 5 x − 26

y = x /5 − 6 /5

Step-by-step explanation:

4 0

4 years ago