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2 years ago

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You take your car in for repairs. The service station charges $200 for parts plus $50 per hour for labor. What is the equation t

hat models the total cost charged by the service station? y= X+

1 answer:

KATRIN_1 [288]2 years ago

6 0

Answer:

y=50x+200 can i get the brainliest?

Step-by-step explanation:

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X-4[x-2(x+6)]=5x+3<br><img src="https://tex.z-dn.net/?f=x%20-%204%20%20%20%5B%20%20x%20-%202%28x%20%2B%206%29%5D%20%20%3D%205x%2

pshichka [43]

The equivalence of the equation is 5x + 48 = 5x + 3.

Since 48 cannot be equal to 3, hence there are no solution.

<h3>What is the value of x?</h3>

Given the equation; x-4[ x - 2( x + 6) ] = 5x + 3

We remove the parentheses

x-4[ x - 2( x + 6) ] = 5x + 3

x-4[ x - 2x - 12 ] = 5x + 3

x - 4x + 8x + 48 = 5x + 3

5x + 48 = 5x + 3

We can go further and collect like terms

5x - 5x + 48 = 3

48 ≠ 3

Since 48 cannot be equal to 3, hence there are no solution.

Learn more about equations here: brainly.com/question/14686792

#SPJ1

7 0

1 year ago

-377 = x - 1,000 i dont under stand it

Naddika [18.5K]

Answer:

$\boxed {x = 623}$

Step-by-step explanation:

Solve for the value of $x$ :

$-377 = x - 1,000$

-Switch sides:

$-377 = x - 1,000$

$x - 1,000 = -377$

-Add $1,000$ on both sides:

$x - 1,000 + 1,000 = -377 + 1,000$

$\boxed {x = 623}$

Therefore, the value of $x$ is $623$ .

4 0

2 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

2 years ago

Help me i need it ASAP

Snowcat [4.5K]

Answer:

m = 13

Step-by-step explanation:

To isolate the variable, you undo what is being done to it. Here, we have the variable being multiplied by 7, hence, we divide both sides by 7 and the result is the variable isolated and solved.

1. 7m = 91

2. (7m)/7 = 91/7

3. m = 13

Hope this helps! (Please consider giving brainliest)

6 0

2 years ago

Express the function h(x)=(x+4)^5 in the form of f of g ( f o g). If f(x)=x^5 , find the function of g(x).

Bess [88]

Answer:

I DONT KNOW DO YOU WANT TO BE MY FRIEND

Step-by-step explanation:

3 0

2 years ago