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3 years ago

What is the solution to the following system of equations?

Mathematics

1 answer:

Semmy [17]3 years ago

3 0

Answer:

(3,2)

Step-by-step explanation:

The system is

x + y = 5

x -y = 1

Add the equations to eliminate y

2x = 6----->x=3

Substitute this value in any equation

3+y = 5----->y=2

The solution is (3,2)

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You are making 2 cakes for your siblings. You want to make sure both cakes have the same

MrRissso [65]

Answer: GOOBLE GOBBLE

Step-by-step explanation: :P

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3 years ago

A rectangle measures 2 <br> 1/2<br> inches by 1 <br> 5/16<br> inches. What is its area?

Norma-Jean [14]

Answer:

it would be 3.28125

Step-by-step explanation:

do base times height to get the area

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3 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

3 years ago

At the dealership where she works, Sade fulfilled 3/10 of her quarterly sales goal in January and another 1/10 of her sales goal

skelet666 [1.2K]

Hi There!

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Problem #1:

At the dealership where she works, Sade fulfilled 3/10 of her quarterly sales goal in January and another 1/10 of her sales goal in February. What fraction of her quarterly sales goal had Sade reached by the end of February?

3/10 + 1/10 = 4/10

4/10 of her goal.

----------------------------------------

Problem #2:

Shane has been monitoring his mileage. According to last week's driving log, he drove 1/10 of a mile in his car and 9/10 of a mile in his truck. How far did Shane drive last week in all?

1/10 + 9/10 = 10/10 = 1

Shane drove 1 mile.

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Problem #3:

For a class experiment, Vina's class weighed a log before and after subjecting it to termites. Before subjecting it to termites, the log weighed 7/10 of a pound. After the termites, the log weighed 3/10 of a pound. How much weight did the termites take from the log?

7/10 - 3/10 = 4/10

The termites took 4/10 pound away from the log.

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Hope This Helps :)

6 0

3 years ago

8x^3+35x^2-31x-9/x+5

BigorU [14]

Answer:

$\frac{8x^{4}+35x^{3}-31x^{2}+5x-9}{x}$

Step-by-step explanation:

6 0

3 years ago