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3 years ago

G(t)=t^3-5t^2 h(t)=-t+5 find (g/h)(t) Please help 50pts who ever gets it

Mathematics

2 answers:

trapecia [35]3 years ago

7 0

Answer:

Step-by-step explanation:

(g/h)(t) = g(t) / h(t)

=( t³-5t² ) /(t+5) we cannot simplify this further so

(g/h)(t) = ( t³-5t² ) /(t+5)

daser333 [38]3 years ago

5 0

Answer:

The awnser is C and trust me

Step-by-step explanation:

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Please show work i rlly need this :)

makvit [3.9K]

Answer:

10/16 (written as a fraction)

Step-by-step explanation:

Starting point

-(-2x+1)=9-14x

Solve the minus sign

2x-1=9-14x

Swap the Xs and the numbers, changing sign

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Do calculations

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Result

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3 years ago

Convert 76,000 ml into L. Please show your work.

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3 years ago

If f(x)=2x^2+(1000/x), find the average rate of change of f(x) from x=a to x=a+h.

galina1969 [7]

Answer:

$\frac{\frac{1000}{x+h}-\frac{1000}{x}}{h}$ is your average rate of change,

Step-by-step explanation:

average rate of change is

$\frac{f(a+h)-f(a)}{a+h-a}$ , by slope formula

simplify this to get $\frac{f(a+h)-f(a)}{h}$ , which is the definition of the derivative as h goes to 0

$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$

since you defined x=a, we can substitute a for x and vice versa to find our derivative.

$\lim_{h \to 0} \frac{(2x^2+\frac{1000}{x+h})-(2x^2+\frac{1000}{x})}{h}$

simplifying

$\lim_{h \to 0} \frac{\frac{1000}{x+h}-\frac{1000}{x}}{h}$ (your average rate of change)

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