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3 years ago

6

Jackson went outside to look at the stars at 8:40 in the evening he went back inside at 9:25 how long did Jackson look at the st

ars?

1 answer:

VashaNatasha [74]3 years ago

8 0

Appreciate it if you mark me brainliest!

Answer:

Jackson looked at the stars for 45 minutes

Good luck!

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Irina-Kira [14]

16 rounded to the tens place = 20
6216 rounded to the thousands place = 6000

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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:

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

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The answer is 24.5
Hope this helps :)

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Read 2 more answers