experts, geniuses, aces and moderators .. need help on the attached. will give brainliest!!! find the derivative of e^x

Mathematics

1 answer:

STatiana [176]3 years ago

4 0

To find the derivative, you must use the chain rule.

If u=x^3+2x:
dy/dx=(dy/du)(du/dx)
dy/du=d/du(e^u)=e^u=e^(x^3 + 2x)
du/dx =d/dx (x^3+2x) = 3x^2 + 2

So dy/dx=
e^(x^3+2x) * (3x^2+ 2)

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1) y=-3x + 4 y = 3x - 2 (1,1)

puteri [66]

Answer:

y=3x-2

Step-by-step explanation:

your question didn't have any instructions so this might not be what you're looking for

set them equal to each other

-3x+4=3x-2

-3x-3x=-2-4

-6x=-6

x=1

plug in x=1

y=3(1)+4 =7

y=3(1)-2 =1

so, when x=1, y=1 in the expression y=3x-2

7 0

2 years ago

In order for this ratio of volumes to be true, what measurements would have to be equal in all 3 solids?

Neporo4naja [7]

The measurements of the figures that should be equal in order for the ratio of volumes between the volumes of the cone, sphere, and cylinder of 1:2:3 to be true is; The radius and the height

<h3>What is a ratio?</h3>

A ratio is a comparison between the magnitudes or measurements of items, expressing the number of times one item in the ratio is contained within another item in the ratio.

The formula for the volume of a cone, $V_p$ is, $V_p$ = (1/3)·π·r₁²·h₁

The formula for the volume of a sphere, $V_s$ , is $V_s$ = (4/3)·π·r₂³

The formula for the volume of a cylinder, $V_c$ , is $V_c$ = π·r₃²·h₃

The ratio of the volumes in the question is 1:2:3

Therefore; The volume of the sphere is twice the volume of the cone, and the volume of the cylinder is three times the volume of the cone, which indicates;

(4/3)·π·r₂³ = 2 × (1/3)·π·r₁²·h₁...(1)

π·r₃²·h₃ = 3 × (1/3)·π·r₁²·h₁...(2)

Equation (1) indicates that we have;

(4/3)·π·r₂³ = 2 × (1/3)·π·r₁²·h₁

Where, h₁ = 2·r₂, we get;

(4/3)·π·r₂³ = 2 × (1/3)·π·r₁²·(2·r₂) = (4/3)·π·r₁²·r₂

Therefore; r₁ = r₂, and the diameter of the sphere is equal to the height of the cone

The radius of the cone is equal to the radius of the sphere

Equation (2) indicates;

π·r₃²·h₃ = 3 × (1/3)·π·r₁²·h₁ = π·r₁²·h₁

Therefore;

r₃²·h₃ = r₁²·h₁