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2 years ago

Which is a solution to the equation y=2x-5?

Mathematics

2 answers:

erma4kov [3.2K]2 years ago

7 0

Answer:

C. (1, -3)

Step-by-step explanation:

Plug in the x to get the y in the equation. The point must fit the formula.

Plug in C. (1, -3), in which x = 1, y = -3:

(-3) = 2(1) - 5

Simplify:

(-3) = 2(1) - 5

-3 = 2 - 5

-3 = -3 (True).

C is your answer.

Zigmanuir [339]2 years ago

4 0

your answer would be c) 1,-3

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2 years ago

Dy/dx if y = Ln (2x3 + 3x).

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Answer:

$\frac{6x^2+3}{2x^3+3x}$

Step-by-step explanation:

You need to apply the chain rule here.

There are few other requirements:

You will need to know how to differentiate $\ln(u)$ .

You will need to know how to differentiate polynomials as well.

So here are some rules we will be applying:

Assume $u=u(x) \text{ and } v=v(x)$

$\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}$

$\text{ power rule } \frac{d}{dx}x^n=nx^{n-1}$

$\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}$

$\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}$

Those appear to be really all we need.

Let's do it:

$\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}$

I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.

Your answer is $\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}$ .

3 0

3 years ago