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3 years ago

A triangle has two sided that are perpendicular, could the triangle be isosceles, equilateral, or scalene?

Mathematics

1 answer:

Sati [7]3 years ago

5 0

The triangle can be isosceles or scalene but not equilateral.

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A straight line is drawn and divided into two angles. One angle is 75°. What is the measurement of the unknown angle?

Artyom0805 [142]

Answer:

105

Step-by-step explanation:

straight line would be 180 degrees

subtract 180-75=105 ^_^

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3 years ago

Please I don’t get this show work please

dlinn [17]

Answer:

8x=p

Step-by-step explanation:

So 8 time the number of hours Carmen works which I used x. Then you’ll get the total.

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2 years ago

Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$

$S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...$

$S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2) + \dfrac{3}{9}(2^2) -...$

$S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...$

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

6 0

3 years ago

What is p−m ÷ p−n equal to? (1 point) a p−m − (−n) b p−m − n c p−m(−n) d p−m ÷ n

mr_godi [17]

Answer:

Please check the formatting in the problem.

Step-by-step explanation:

As written, "p−m ÷ p−n" is only p - m/p - n. None of the answer options comes close. Is the expression witten correctly?

The Answer Options are also unusual:

a) p−m − (−n) is just p-m+n

b) p−m − n

c) p−m(−n) is just p +mn

d) p−m ÷ n is p = (m/n)

5 0

2 years ago

12. Three fifths of the children at my party like chocolate ice cream. How many of the 20 children at my party do not like choco

vichka [17]

For number 12 - 0.63

5 0

2 years ago