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Rasek [7]
4 years ago
13

Jill, Mateo, and Rasheed each have some stickers.

Mathematics
1 answer:
navik [9.2K]4 years ago
3 0

Answer:

430+110-90

Step-by-step explanation:

the second option

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

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