2 years ago

Classify each function as linear or nonlinear.

Mathematics

1 answer:

faltersainse [42]2 years ago

5 0

4x-y=3 linear,2x²=12-8y linear,-6x=3y -17..linear

You might be interested in

What does a positive correlation tell you about the graph that compares advertising costs and sales. Would the owners of the com

MArishka [77]

A positive correlation means sales are rising because of advertising
the steeper the chart the more money is being made, so the owner would be happier with a steeper slope

5 0

2 years ago

Read 2 more answers

Byron has 1 7/10 kilograms of black pepper. He uses 7/8 of the black pepper and splits between 7 pepper shakers. How much pepper

IrinaVladis [17]

Amount of black pepper = 1 7/10 kg = 17/10 kg.
Amount used is
(7/8)*(17/10) = 119/80 kg

This amount is split between 7 shakers. Each shaker has
(119/80)/7 = 17/80 kg = 0.2125 kg

Answer: 17/80 kg (or 0.2125 kg)

4 0

1 year ago

jillian is trying to save water, so she reduces the size of her square grass lawn by 8 feet on each side. the area of the smalle

8_murik_8 [283]

We can solve this equation by using the Square Root Method.
First, take the square root of each side of the equation:
(x-8)^2 = 144 becomes x-8 = 12
Then add 8 to both sides.
x=20

The original lawn was d. 20 feet by 20 feet.

5 0

2 years ago

Read 2 more answers

The following table shows the time in various cities when it is 4:00 a.m. in New York.

Sati [7]

It's Dublin - San Juan

Departure Friday at 1:30 Pm (Dublin Time) ==> Time San Juan 9:30 (-4 Hours)

Travel time 16 Hrs ( Dub Sat at 6:30PM) & arriving San JUan on Sat at 9:30 +16;30 26 that means 2 AM (26-24)

8 0

2 years ago

Read 2 more answers

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

monitta

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

We are required to solve the equation for P(t).

$\dfrac{dP}{dt}=kM-kP(t)\\$Add kP(t) to both sides\\\dfrac{dP}{dt}+kP(t)=kM\\$Taking the integrating factor\\e^{\int k dt} =e^{kt}\\$Multiply all through by the integrating factor\\\dfrac{dP}{dt}e^{kt}+kP(t)e^{kt}=kMe^{kt}\\\dfrac{dP}{dt}e^{kt}=kMe^{kt}\\(Pe^{kt})'=kMe^{kt} dt\\$Take the integral of both sides with respect to t\\\int (Pe^{kt})'=\int kMe^{kt} dt\\Pe^{kt}=kM \int e^{kt} dt\\Pe^{kt}=\dfrac{kM}{k} e^{kt} + C_0, C_0$ a constant of integration$

$Pe^{kt}=Me^{kt} + C\\$Divide both side by e^{kt}\\P(t)=M+Ce^{-kt}\\P(t)=M+Ce^{-kt}\\$Therefore:\\P(t)=M+Ce^{-kt}$

4 0

2 years ago