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frez [133]
3 years ago
10

Find the derivative g(x)=ln(2x^2+1)

Mathematics
1 answer:
lana [24]3 years ago
8 0

Answer:

4x / (2x^2+1)

Step-by-step explanation:

g(x)=ln(2x^2+1)

Using u substitution

u= 2x^2 +1

du = 4x  dx

g'(x) = d/du ( g(u) du)

We know that the derivative of ln(u) = 1/u  since this is always positive

        = 1/ u* du

 Substituting x back into the equation

g'(x) = 1/(2x^2+1) * 4x

       = 4x / (2x^2+1)

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At what angle do the angle bisectors of two same side interior angles intersect in construction with two parallel lines and a tr
jeka94
Same-side interior angles are a pair of angles on one side of a transversal line, and on the inside of the two parallel lines being intersected.

The sum of same-side interior angles is equal to 180 degrees.

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Parallel lines are lines that run alongside each other that never intersect.

An angle bisector is a line that divides an angle into two equal parts.

Given that the sum of same side interior angles is equal to 180 degrees, then the sum of the angle formed by the angle bisector is equal to 90 degrees.
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3 0
3 years ago
How many solutions does the equation −4a + 4a + 3 = 8 have?
vladimir1956 [14]

Answer:

none

Step-by-step explanation:

7 0
3 years ago
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erastova [34]
What are you trying to ask here?
3 0
3 years ago
Given: ΔPSQ, PS = SQ
horrorfan [7]
<h2>Answer:</h2>

To solve this problem we will use Heron's formula:

A=\sqrt{s(s-a)(s-b)(s-c)}

Where a, \ b \ and \ c are the side lengths of the triangle and s is the semiperimeter (half the perimeter of the triangle). We know that:

Perimeter \ P=\triangle PSQ=PS+PQ+SQ: \\ \\ \triangle PSQ=P=50 \\ \\ Semiperimeter \ s: \\ \\ s=\frac{P}{2}=25

Also:

(I) \ PS=SQ \\ \\ (II) \ SQ-PQ = 1 \\ \\ (III) \ PS+PQ+SQ=50 \\ \\ \\ (I) \ into \ (III): \\ \\ SQ+PQ+SQ=50 \\ \\ \therefore (IV) \ 2SQ+PQ=50 \\ \\ From \ (II): \\ \\ PQ=SQ-1 \\ \\ (II) \ into \ (IV): \\ \\ 2SQ+(SQ-1)=50 \\ 3SQ-1=50 \\ 3SQ=51 \\ \\ \boxed{SQ=17} \\ \\ \boxed{PS=17} \\ \\ PQ=SQ-1=17-1 \therefore \boxed{PQ=16}

Finally:

A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ A=\sqrt{s(s-PS)(s-SQ)(s-PQ)} \\ \\ A=\sqrt{s(s-17)(s-17)(s-16)} \\ \\ A=\sqrt{25(25-17)(25-17)(25-16)} \\ \\ \boxed{A=120}

7 0
3 years ago
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