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Flura [38]
2 years ago
6

HELP PLS THIS IS HARD WLH

Mathematics
1 answer:
AnnyKZ [126]2 years ago
8 0

Answer:

D

Step-by-step explanation:

A'(-2*3/2=-3,3*3/2=4.5), B'(-6,-6), C'(6,0), D'(7.5,6)

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Find the derivative of the function by using the Product Rule. Simplify your answer. f(x) = (x - 3)(x + 3)
Morgarella [4.7K]

Answer:

f'(x)=2x

Step-by-step explanation:

Given : Function f(x)=(x-3)(x+3)

To find : The derivative of the function by using the Product Rule ?

Solution :

The product rule of derivative is

\frac{d}{dx}(u\cdot v)=uv'+vu'

Here, u=x-3 and v=x+3

\frac{d}{dx}((x-3)\cdot (x+3))=(x-3)\frac{d}{dx}(x+3)+(x+3)\frac{d}{dx}(x-3)

\frac{d}{dx}((x-3)\cdot (x+3))=(x-3)1+(x+3)1

\frac{d}{dx}((x-3)\cdot (x+3))=x-3+x+3

\frac{d}{dx}((x-3)\cdot (x+3))=2x

Therefore, f'(x)=2x

3 0
3 years ago
Read 2 more answers
Find an equation for the line with slope m through the point (a,c).
lidiya [134]
If you want the answer in point slope form then,

y-y1 = m(x-x1)
y-c = m(x-a)

---------------------------------------

If you want the answer in slope intercept form, then solve for y

y-c = m(x-a)
y-c = mx-ma
y-c+c = mx-ma+c
y = mx-ma+c
y = mx+c-ma
y = mx+(c-ma)

For this answer in slope intercept form the slope is m and the y intercept is c-ma

---------------------------------------

If you want the answer in standard form, then get the variable terms to the left side. Have the constant terms on the right side.

y = mx+c-ma
y-mx = mx+c-ma-mx
-mx+y = c-ma

Optionally you can multiply both sides by -1 to get mx-y = -c+ma but it will depend on your book if this step is carried out or not.

4 0
3 years ago
HELP PLEASE !!!
jasenka [17]

Answer:

The owner should gather more.

Step-by-step explanation:

He should gather more so he can make sure he is correct.

3 0
2 years ago
Read 2 more answers
-2x^4 +50x^2+0x-3/x-5
Gemiola [76]
1) "x2" was replaced by "x^2". 1 more similar replacement(s).
Raising zero to a power is not allowed
3 0
3 years ago
Y=2x-3 (-5,3) for geometry
Doss [256]
Y=2x+159 that's the answer .
8 0
3 years ago
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