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3 years ago

You can ignore a and b but can someone help me with c and d plz :)

Mathematics

2 answers:

katrin2010 [14]3 years ago

7 0

Answer:

(below)

Step-by-step explanation:

c) the angles are in the interior on the same side of the transversal, making them supplementary

equation: 2x + 20 + 120=180 and then when solved you would get x =20

d) the angles are corresponding, which makes them congruent

equation: 3x + 10 = 100 and then when solved you would get x = 30

garri49 [273]3 years ago

7 0

For C, the answer is x = 20. Explanation: the sum of the two angles is equal to 180 degrees, since they are across parallel lines. Plug the two angles into an equation that equals 180. You should have (2x + 20) + 120 = 180. Simplify the equation by subtracting 120 from both sides. You should now have 2x + 20 = 60. Subtract 20 from both sides. 2x = 40. Now simplify the equation by dividing both sides by 2, and you get your answer, x = 20. The answer for D is x = 30. Explanation: Since the angles are on the same side of parallel lines, they are equal. Plug the angles into an equal equation: 3x + 10 = 100. Subtract 10 from both sides to get 3x = 90. Now divide both sides by 3 to get your answer, x = 30.

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3 years ago

A sector of a circle has an arc length of pi cm and a central angle of pi over 6 radians. What is the area of the sector?

liberstina [14]

Answer:

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Step-by-step explanation:

Mathematically pi radians is 180 degrees

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2 years ago

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2 years ago

Please, I need help in this ??

nignag [31]

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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3 years ago