All categories

2 years ago

System of Equations: Solve for (x,y) given the system of linear equations "X 1 point

Mathematics

1 answer:

Elenna [48]2 years ago

8 0

Answer:

(x, y) = (4, 0)

Step-by-step explanation:

It looks like you want to solve ...

x - y = 4
x + y = 4

Adding the two equations gives ...

2x = 8

x = 4 . . . . divide by 2

Then ...

y = 4 - x = 0

The solution is (x, y) = (4, 0).

_____

If you think about what you're seeing when you read the equations, you realize that adding or subtracting y gives the same result. Hence y=0 and x=4.

You might be interested in

Differentiate x^2 e^x logx

zimovet [89]

Product rule:

$\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)$

$=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}$

Power rule:

$\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x$

The exponential function is its own derivative:

$\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x$

Assuming the base of $\log x$ is $e$ , its derivative is

$\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x$

But if you mean a logarithm of arbitrary base $b$ , we have

$y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}$

$\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$

8 0

3 years ago

What is the surface area of the right prism given below?

Sloan [31]

Answer:

10*15= 150

150 *8= 1,200

1,200 ÷ 2 = 600.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can someone do this for me?

EastWind [94]

Do you know your places? Like tenth place?

6 0

3 years ago

Answer pls i need help

nataly862011 [7]

Answer:

What/where is the question you have for me?

Step-by-step explanation:

5 0

3 years ago

A data set is made up of these values.

marissa [1.9K]

I think that the answe my be b

6 0

3 years ago

Read 2 more answers