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3 years ago

Is the statement (3^5)^4 equal to (3^4)^5 ? explain the reasoning, DUE TOMORROW

Mathematics

2 answers:

Shkiper50 [21]3 years ago

4 0

Note that both 5 and 4 are exponents, and their product is 5*4=20, or 4*5=20. Thus,

(3^5)^4 is equal to (3^4)^5

Review:

(a to the power b)to the power c = a to the power bc.

Thus, (3 to the power 5)^4 = 3^20; the same is true if you reverse the positions of the 5 and the 4.

Juli2301 [7.4K]3 years ago

4 0

Yes, because they are basically the same problem since you have to multiply the exponents together.

(3^4)^5 = (3^4*5) = 3^20 = 3486784401

(3^5)^4 = (3^5*4) = 3^20 = 3486784401

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Simplify the sum. (8u^3+8u^2+6)+(4u^3-6u+3)

rewona [7]

Answer:

$\boxed{12u^3+8u^2-6u+9}$

Step-by-step explanation:

$(8u^3+8u^2+6)+(4u^3-6u+3) = 8u^3+8u^2+6+4u^3-6u+3$

Once

$8x+4x = 12x, \text{ as } x = u^3$

$8u^3+8u^2+6+4u^3-6u+3 = 12u^3+8u^2+6-6u+3 = \boxed{12u^3+8u^2-6u+9}$

You don't need to factor.

7 0

3 years ago

Given that y = sin(x+y),find the derivative when (x,y)=(π,0)

lisov135 [29]

<h2>Answer:</h2>

Shown in the explanation

<h2>Step-by-step explanation:</h2>

Recall that an implicit function is a relation given by the form:

$R(x_{1},\ldots, x_{n})=0$

Where $R$ is a function of two or more variables. In this case, that function is:

$y = sin(x+y)$

and is implicit because we can define it as:

$y-sin(x+y)=0$ having two variables.

So, let's take the derivative:

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\sin \left(x+y\right)\right) \\ \\$

Applying chain rule:

$\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)$

But:

$\frac{d}{dx}\left(y\right)=y'$

Therefore:

$y'=\cos \left(x+y\right)\left(1+y'\right)$

Isolating $y'$ :

$\frac{d}{dx}\left(y\right)=y'=\frac{\cos \left(x+y\right)}{1-\cos \left(x+y\right)}$

When $(x,y)=(\pi,0)$ :

$\frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi+0\right)}{1-\cos \left(\pi+0\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi\right)}{1-\cos \left(\pi\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{-1}{1-(-1)} \\ \\ \boxed{\frac{d}{dx}\left(y\right)|_{(\pi,0)}=-\frac{1}{2}}$