PLEASE HELP

2 answers:

7 0

Answer: I believe that the answer for this is that she is going over budget considering that 100$ in debt she won't be able to pay because without the debt she would have to pay 1,605 out of her 1,700 so add 100 to 1,605 and she would be over budget

Step-by-step explanation:

andre [41]3 years ago

3 0

I believe that the answer for this is that she is going over budget considering that 100$ in debt she won't be able to pay because without the debt she would have to pay 1,605 out of her 1,700 so add 100 to 1,605 and she would be over budget

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identify the type of function that models the labor cost of repairing a computer if the charge is $50 per hour or fraction there

mezya [45]

Answer: Linear

<u>Step-by-step explanation:</u>

Labor Cost = $50 per hour → y = 50x

Since x has an exponent of 1, it is a linear function.

Bonus:

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exponent of 4: quartic function

5 0

3 years ago

Heyyy!!! How was yall's day

Serjik [45]

Good my teacher wasn’t going overboard today

5 0

2 years ago

I NEED HELP ON THIS PROBLEM!!! PLEASE

Allushta [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$5x+8y=-9\implies 8y=-5x-9\implies y=\cfrac{-5x-9}{8} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{5}{8}} x-9\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

well then, so since this equation has that slope therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{-5}\implies \cfrac{8}{5}}}$

so we're really looking for the equation of a line whose slope is 8/5 and runs through (10,10)

$(\stackrel{x_1}{10}~,~\stackrel{y_1}{10}) ~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{8}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{\cfrac{8}{5}}(x-\stackrel{x_1}{10})$