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3 years ago

15

a car repair shop charge $80 per hour for labor and $210 for the parts of the car if it's auto parts of the repair for the car w

as 490 how many hours did it take to repair the car

1 answer:

juin [17]3 years ago

7 0

490-210= 280; 280/80= 3.5 or 3.5 hours

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for the school play four rows of chairs are set up there are 22 chairs in each row how many chairs are in all

allsm [11]

There are about 88 chairs in all.
All you have to do is 22x4 which equals 88. :)

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2 years ago

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PLEASE HELP ME FAST I NEED TO FINISH THIS!! ( It's about linear equation from a slope and y- intercept)

alex41 [277]

Y=Mx+b OR y= slopex+ y intercept

y=9x-8

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2 years ago

which of the folowing could be the equation for a parabola that opens to the left with vertex (-17, 2)

KATRIN_1 [288]

Answer:

x = a(y+17)^2 + 2 where a < 0

Step-by-step explanation:

Since the parabola is horizontal, the formula of the parabola is x = a(y-h)^2 + k where (h,k) is the vertex.

Since it opens to the left, a must be negative. The vertex is (-17, 2), so h = -17 and k = 2.

Substituting these in, we find that the equation is

x = a(y+17)^2 + 2 where a is a real number that is less than 0 (i.e. negative)

I hope this helps! :)

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2 years ago

42,199,279 rounded to the nearest ten million

Simora [160]

Answer:

40,000,000

42,199,279

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3 years ago

A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair a

vesna_86 [32]

$Chair=x$

$Table=y$

$Desk=z$

$\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}$

keep the first row as normal, then in the other ones, we can isolate Y and X

$\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace at first row...

$\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace the Z to discovery the other value

$\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}$

$\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}$

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3 years ago