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DanielleElmas [232]

2 years ago

12

Which of the following operations could you perform on both sides of the given equation to solve it? Check all that apply. 8x -

6 = 2x + 24

1 answer:

rodikova [14]2 years ago

7 0

It equals 26x because you add the 24 to the 2x

You might be interested in

For the following cylinder, what is the area of the two bases in square centimeters?

Bezzdna [24]

Cylinder formula is 2pi(r)^2 + 2pi(r)h

2pi(0.25)^2 + 2pi(0.25)(1)

Chuck that in calculator and evaluate

8 0

3 years ago

A sphere has a diameter of 14.5 inches.

dlinn [17]

Answer:

≈ 660.5 in²

Step-by-step explanation:

The surface area (A) of a sphere is calculated using the formula

A = 4πr² ← r is the radius

Given diameter = 14. 5 then radius = 7.25, thus

A = 4π × 7.25² = 4π × 52.5625 ≈ 660.5 in²

4 0

3 years ago

Mr. Johnson bought 22 cupcakes and 30 drinks for his class. Ms. Richards bought 30 cupcakes and 35 drinks for her class. Mr. Joh

Lostsunrise [7]

Cupcakes =c
Drinks = d
22 c + 30d = $44.40...eqn 1
30c + 35d = $53.75...eqn 2
Multiply eqn 1 by 7 and eqn 2 by 6
154c +210d =$310.80...eqn 3
180c+ 210d = $322.5...eqn 4
Eqn 4 - eqn 3
26 c = $11.70
C = 11.7/26 = 45c
Subst c = 45c in eqn 1
$9.90 + 30d =$44.40
d = ($44.40-$9.90)/30 = $1.15

4 0

3 years ago

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

NeTakaya

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

The graph given above shows the following function.

serg [7]

Answer:

I think its A

Step-by-step explanation:

8 0

2 years ago