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3 years ago

Given the system of equations, what is the solution? x + 2y = 11 x - 2y = -1

Mathematics

1 answer:

zhenek [66]3 years ago

4 0

We need to solve x+2y=11 for x

Let's start by adding -2y to both sides

x+2y -2y = 11-2y

x= -2y+11

Now substitute -2y+11 for x in x-2y =-1

x-2y=-1

-2y+11-2y=-1

-4y+11=-1

Add -11 to both sides

-4y+11-11=-1-11

-4y=-12

Divide both sides by -4

-4y/-4 = -12/-4

y= 3

Substitute 3 for y in x=-2y+11

x= -2y+11

x = (-2)(3)+11

x=-6+11

x=5

The answer is x=5 and y=3

(5,3)

I hope that's help:)

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I need help proving this ASAP

Ket [755]

Answer:

See explanation

Step-by-step explanation:

We want to show that:

$\tan(x + \frac{3\pi}{2} ) = - \cot \: x$

One way is to use the basic double angle formula:

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ \sin(x) \cos( \frac{3\pi}{2} ) + \cos(x) \sin( \frac{3\pi}{2}) }{\cos(x) \cos( \frac{3\pi}{2} ) - \sin(x) \sin( \frac{3\pi}{2}) }$

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ \sin(x) ( 0) + \cos(x) ( - 1) }{\cos(x) (0) - \sin(x) ( - 1) }$

We simplify further to get:

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ 0 - \cos(x) }{0 + \sin(x) }$

We simplify again to get;

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{- \cos(x) }{ \sin(x) }$

This finally gives:

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = - \cot(x)$

6 0

3 years ago

A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po

Mariana [72]

Answer:

The work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

I hope it helps you!

6 0

2 years ago

Calculate the volume of the triangular pyramid below.

eduard

Answer:

Step-by-step explanation:

1/2x10x3.6=18

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8 0

2 years ago

Suppose you know that the volume of the following prism is represent by V(x) = -2x^3 + 14x^2 + 120x.

erastova [34]

Step-by-step explanation:

-2x³ + 14x² + 120x

= -2x(x² - 7x - 60)

= -2x(x - 12)(x + 5).

The other 2 dimensions are -2x and (x + 5).

When using a graphing calculator to maximise the volume of the prism, we get x = 7.378. However this value is not reasonable as it makes -2x and (x - 12) negative, which are the sides of the prism.

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2 years ago

K(a)= 4a-4, Find k(-9)

FromTheMoon [43]

We simply replace a with -9

k(-9) = 4 * -9 - 4

k(-9) = -40

5 0

2 years ago