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OverLord2011 [107]
2 years ago
6

Consider triangle ABC with the measure of an angle B = 60゚ and sidelinks a equal 4 &c equal 5 what option listed is an expre

ssion that is equivalent to the length of side B
Mathematics
1 answer:
ivann1987 [24]2 years ago
8 0

The law of cosine helps us to know the third side of a triangle when two sides of the triangle are already known the angle opposite to the third side is given. The correct option is B.

<h3>What is the Law of Cosine?</h3>

The law of cosine helps us to know the third side of a triangle when two sides of the triangle are already known the angle opposite to the third side is given. It is given by the formula,

c =\sqrt{a^2 + b^2 -2ab\cdot Cos\theta}

where

c is the third side of the triangle

a and b are the other two sides of the triangle,

and θ is the angle opposite to the third side, therefore, opposite to side c.

The length of the sidelink b using the cosine rule can be written as,

b = \sqrt{a^2+c^2-2ac\cos(\angle B)}\\\\b = \sqrt{4^2+5^2 - 2(4)(5)(\cos 60^o)}\\\\b = \sqrt{16+25-20}

Hence, the correct option is B.

The complete question is:

Consider ABC with the measure of angle B equal to 60 degrees, and side lengths a=4 and c=5. Which option lists an expression that is equivalent to the length of side b?

Options are given in the image below.

Learn more about the Law of Cosine:

brainly.com/question/17289163

#SPJ1

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Answer:b

Step-by-step explanation

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Please, I need help in this ??
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Answer:

\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c

Step-by-step explanation:

\int\frac{x^{4}}{x^{4} -1}dx

Adding and Subtracting 1 to the Numerator

\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx

Dividing Numerator seperately by x^{4} - 1

\int 1 + \frac{1}{x^{4}-1 }\, dx

Here integral of 1 is x +c1 (where c1 is constant of integration

x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx----------------------------------(1)

We apply method of partial fractions to perform the integral

\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}------------------------------------------(2)

\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}

1 = A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)-------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = \frac{1}{4}

1 = B(-1-1)(1+1)

B = -\frac{1}{4}

1 = C(i-1)(i+1)

C = -\frac{1}{2}

Substituting A, B, C in equation (2)

\int\frac{x^{4}}{x^{4} -1}dx = \int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }

On integration

Here \int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx

\int\frac{x^{4}}{x^{4} -1}dx = \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2

Here c1 + c2 can be added to another and written as c

Therefore,

\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c

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Step-by-step explanation:

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