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Anastasy [175]
2 years ago
10

Need some help with the questions in the image​

Mathematics
1 answer:
ollegr [7]2 years ago
3 0

Answer:

1a. E = IR, I = 5, R = 3

E = 5 × 3

E = 15

b. A = πrl, π = 3.14, r = 5, l = 20

A = 3.14 × 5 × 20

A = 314

c. S = 90 ( 2n - 4 ), n = 6

S = 90 ((2 × 6) - 4)

S = 90 ( 12 - 4 )

S = 90 ( 8 )

S = 720

You might be interested in
Translate the phrase into a numerical expression.
Alexxandr [17]
1. Seven multiplied by three
7*3 the little star mean times

2. Twice the opposite of eleven= 2x+11
22 less than 3 times the number
3x-22
3x-22=2x+11
3x-2x=11+22
x= 33
So , it would be 2x+11 x is 33

Hope this helps good luck
4 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
Debra purchased a prepaid phone card for $25 . long distance calls cost 11 cents a minute using this card. debra used her card o
alex41 [277]
Amount used = $25 - $23.35 = $1.65

Number of minutes used = $1.65 ÷ $0.11 = 15 mins

Answer: 15 mins
6 0
4 years ago
The price of a coffee was $2.15 yesterday. Today, the price rose to $2.30. Find the percentage increase. Round your answer to th
Phantasy [73]
7.5%

Hope this helped
8 0
4 years ago
Describe fully the single transformation that maps triangle A into triangle B
Kitty [74]

Answer:

Reflection over the line y=-1

Step-by-step explanation:

5 0
3 years ago
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