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natka813 [3]
3 years ago
10

NEED HELP PLEASE AND THANKS!

Mathematics
1 answer:
marta [7]3 years ago
5 0

Answer:

The answer is A.

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Determine the domain for the graphed function:<br> Please help meeee
Rom4ik [11]

Answer:

D

Step-by-step explanation:

First, domain refers to the x axis. So you would find the lowest number on the X axis, -2, and then you look at the kind of dot on that number. It is an open dot, which means that it is all the numbers up to -2, but does not include -2. Then you find the highest number, in this case 2. Looking at the dot that is marking it, it is a closed dot, meaning it includes the number 2. So the domain would be numbers between -2 and 2, but does not include -2. all numbers greater than -2, x, all numbers less than and equal to 2.

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Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 1
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Answer:

attached

Step-by-step explanation:

attached

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Help please !!<br> The graph of the invertible function f is shown on the grid below.
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Answer:

4

Step-by-step explanation:

→ We go up the y axis to 6 and read of the x coordinate which is 4. This is because an inverse function does the the opposite i.e.  if f ( x ) = y

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What is the slope of the line represented by the equation y= x/6-5?
Alexxandr [17]
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Use integration by parts to derive the following formula from the table of integrals.
emmasim [6.3K]

Answer:

I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)

Step-by-step explanation:

for

I= ∫x^n . e^ax dx

then using integration by parts we can define u and dv such that

I= ∫(x^n) . (e^ax dx) = ∫u . dv

where

u= x^n → du = n*x^(n-1) dx

dv= e^ax  dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)

then we know that

I= ∫u . dv = u*v - ∫v . du + C

( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =

(u*v) - ∫v*du + C )

therefore

I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C

I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)

5 0
3 years ago
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