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3 years ago

Question 3 Please. I'm really stuck :(

Mathematics

2 answers:

Stells [14]3 years ago

7 0

Let ∠RTS=∠RST = a (say)
∠QUA=∠QSU= b(say)
then we know , at point S, a+40+b=180. so, a+b=140 we'll use this later.

consider trianglePQR, ∠P+∠Q+∠R=180
i.e.P+(180-2b)+(180-2a)=180
P+180+180-2(a+b)=180 ⇒P+180-2(a+b)=0 ⇒P=2(140)- 180=280-180=100
hence,answer is E

dem82 [27]3 years ago

4 0

Answer B - 70 degrees
Triangle STU is an isosceles one. Angle s is 40. Find angle TSR (180-40)/2 = 70. Now find angle SRT (180-70)/2 = 55. Being isosceles angle SRT = angle USQ = 55. Now find angle TPU (180-(55+55)) = 70

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irga5000 [103]

Answer:

100%-75%=25%

Step-by-step explanation:

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4 years ago

A customer needs to seed an area 75 feet by 50 feet in size. Each bag of seed can cover 25 square feet of land. How many bags of

Ronch [10]

Answer:

150 bags

Step-by-step explanation:

Given the total area:

The area will be:

=75*50

= 3750 square feet

As it is given that one bag covers 25 square feet. To find the total number of bags we have to find how many 25s will be in 3750 square feet.

So,

Total number of bags = 3750 / 25

= 150 bags

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5 0

3 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

1. Calculate the slope between the two points (2,5) and (-3,-4).

Oksana_A [137]

To find the slope of thine that passes through these points, use the slope formula. It can be read as “slope equals the second y-coordinate minus the first y-coordinate over the second x-coordinate minus the first x-coordinate.”

So we have $\frac{-4 - 5}{-3 - 2}$ and this simplifies to -9/-5 or 9/5.

Remember, a negative divided by a negative is a positive.

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4 years ago

The next model of a sports car will cost 13.8% more than the current model. The current model costs $54,000. How much will the p

garik1379 [7]

I think that the increase would be $7,452, and the new model would cost $61,452. I may be incorrect though.

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3 years ago

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