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TiliK225 [7]
3 years ago
6

What’s the gcf (greatest common factor) of this expression

Mathematics
2 answers:
Firdavs [7]3 years ago
6 0

Answer:

x

Step-by-step explanation:

You go by the LEAST DEGREE TERM POSSIBLE. Now, for coefficients, they do not have any common factors other than 1, so that is out of the question.

I am joyous to assist you anytime.

love history [14]3 years ago
5 0

Answer:

x

Step-by-step explanation:

just look at them what do they have in common?

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Let the region R be the area enclosed by the function f(x)=ln(x) and g(x)= 1/2x-2. Find the volume of the solid generated when t
Neporo4naja [7]

Answer:

V=61.66

Step-by-step explanation:

This problem can be solved by using the expression for the Volume of a solid with the washer method

V=\pi \int \limit_a^b[R(x)^2-r(x)^2]dx

where R and r are the functions f and g respectively (f for the upper bound of the region and r for the lower bound).

Before we have to compute the limits of the integral. We can do that by taking f=g, that is

f(x)=g(x)\\ln(x)=\frac{1}{2}x-2

there are two point of intersection (that have been calculated with a software program as Wolfram alpha, because there is no way to solve analiticaly)

x1=0.14

x2=8.21

and because the revolution is around y=-5 we have

R=ln(x)-(-5)\\r=\frac{1}{2}x-2-(-5)\\

and by replacing in the integral we have

V=\pi \int \limit_{x1}^{x2}[(lnx+5)^2-(\frac{1}{2}x+3)^2]dx\\

V=\pi [28x+\frac{1}{x}+xln^2x-12xlnx-6lnx]  

and by evaluating in the limits we have

V=61.66

Hope this helps

regards

3 0
3 years ago
What is 3.3(x-8)-x=1.2
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Multiply 3.3 by everything in the parenthesis.

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2.3x - 26.4 = 1.2

Add 26.4 to both sides.

2.3x = 27.6

Divide 2.3 on both sides.

x = 12

Hope this helps!
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<span>D) perpendicular bisector <em>I believe.
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Step-by-step explanation:

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Guys pls help me! ASAP!
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Answer:

Step-by-step explanation:

2. area of a triangle is 1/2 b h

1/2 (5 * 2) = 5 yd squared

area of the circle is pi r squared

you only need 1/2 of this area

find all these areas and add them together.

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