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3 years ago

What is 9 more then the sum of a number and three is 35 into a equation

1 answer:

4 0

Alright, first let's make a variable for the number.
'n'
Pretty easy right?
Now, we know that the equation contains the sum of n and 3.
n + 3
Since 35 is 9 more than the sum of n and 3:
(n + 3) + 9 = 35
Great! Now we have the equation!
We can open the parentheses and add 9 and 3
n + 12 = 35
Subtract both sides by 12:
n = 23
The unknown number is 23!

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3 years ago

Find the derivative f (x ) = (x-5)^2 (3-x)^2

zavuch27 [327]

Given:

The function is

$f(x)=(x-5)^2(3-x)^2$

To find:

The derivative of the given function.

Solution:

Chain rule of differentiation:

$[f(g(x))]'=f'(g(x))g'(x)$

Product rule of differentiation:

$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$

We have,

$f(x)=(x-5)^2(3-x)^2$

Differentiate with respect to x.

$f'(x)=(x-5)^2\dfrac{d}{dx}(3-x)^2+(3-x)^2\dfrac{d}{dx}(x-5)^2$

$f'(x)=(x-5)^2[2(3-x)(0-1)]+(3-x)^2[2(x-5)(1-0)]$

$f'(x)=(x^2-10x+25)(-6+2x)+(9-6x+x^2)(2x-10)$

$f'(x)=(x^2)(-6)+(-10x)(-6)+(25)(-6)+(x^2)(2x)-10x(2x)+25(2x)+(9)(2x)+(-6x)(2x)+x^2(2x)+9(-10)+(-6x)(-10)+x^2(-10)$

On further simplification, we get

$f'(x)=-6x^2+60x-150+2x^3-20x^2+50x+18x-12x^2+2x^3-90+60x-10x^2$

$f'(x)=(2x^3+2x^3)+(-6x^2-20x^2-12x^2-10x^2)+(60x+50x+18x+60x)+(-90-150)$

$f'(x)=4x^3-48x^2+188x-240$

Therefore, the derivative of the given function is $f'(x)=4x^3-48x^2+188x-240$ .

3 0

3 years ago

Find m<YDC please help!

AlekseyPX

Answer:

∠YDC measures 140°.

Step-by-step explanation:

We can consider using the Exterior Angle Theorem. According to the theorem, the exterior angle of a triangle is equal to the two opposite interior angles.

In other words:

$\displaystyle m \angle YDC = m\angle DCB + m\angle CBD$