The blue dot is at what value on the number line? 12 8

Mathematics

1 answer:

Kobotan [32]3 years ago

4 0

Answer:

Step-by-step explanation:

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The yearly cost in dollars, y, at a video game arcade based on total game tokens purchased, x, is y = x + 60 for a member and y

morpeh [17]

We have the yearly cost in dollars y at a video game arcade based on total game tokens purchased $x$ . So we know that:

 $y=x+60:\ equation \ for \ a \ member$ 

 $y=x:\ equation \ for \ a \ nonmember$ 

Then we can study this problem by using the graph in the figure below. We know that if there's no any purchase, the yearly cost for a member will be $60 and for a nonmember there will not be any cost. From this, we can affirm that the cost of membership is equal to $60.

On the other hand, both members and nonmembers will pay the same price on the total game tokens purchased, this is true because of the same slope that members and nonmembers have in the equations.

8 0

3 years ago

Read 2 more answers

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$