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Shtirlitz [24]
2 years ago
7

Let f(x) = e ^3x/5x − 2. Find f'(0).

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
4 0

Answer:

Step-by-step explanation:

Our friend asking what the actual function is has a point. I completed this under the assumption that what we have is:

f(x)=\frac{e^{3x}}{5x-2} and used the quotient rule to find the derivative, as follows:

f'(x)=\frac{e^{3x}(5)-[(5x-2)(3e^{3x})]}{(5x-2)^2} and simplifying a bit:

f'(x)=\frac{5e^{3x}-[15xe^{3x}-6e^{3x}]}{(5x-2)^2}and a bit more to:

f'(x)=\frac{5e^{3x}-15xe^{3x}+6e^{3x}}{(5x-2)^2} and combining like terms:

f'(x)=\frac{11e^{3x}-15xe^{3x}}{(5x-2)^2} and factor out the GFC in the numerator to get:

f'(x)=\frac{e^{3x}(11-15x)}{(5x-2)^2}  That's the derivative simplified. If we want f'(0), we sub in 0's for the x's in there and get the value of the derivative at x = 0:

f'(0)=\frac{e^0(11-15(0))}{(5(0)-2)^2} which simplifies to

f'(0)=\frac{11}{4} which translates to

The slope of the function is 11/4 at the point (0, -1/2)

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y = −2x + 13


y = 2x − 3


Set the equations to equal and solve for x:

-2x + 13 = 2x -3

Add 2X to each side:

13 = 4x -3

Add 3 to each side:

16 = 4x

Divide both sides by 4:

x = 16/4

x = 4


Now replace x with 4 in one of the equations and solve for y:


y = -2(4) +13

y = -8 +13

y = 5


x = 4 and y = 5 , which is the ordered pair (4,5)

Both numbers are positive values, so the second graph is correct.

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The answer for 2ab(xa^9+ya^3-6) is 2a^10bx + 2a^4by-6ab.

What is the mathematical expansion?

A mathematical expansion can be described as the finite combination of symbols which can be seen as one having rules which is usually based on the context that is been used.

The solution for the question is 2a^10bx + 2a^4by-6ab

Therefore, the expansion  of 2ab(xa^9+ya^3-6) is 2a^10bx + 2a^4by-6ab.

Learn more about expansion at:

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Step-by-step explanation:

<em>Since the scale has two plates, we can place weights on either side and also the object so it can be balanced. </em>

This is a key part of the problem, it's not only on the other side of the scale, but on both sides.

Let's do the math now.

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