3 years ago

Please answer question a & b.

Mathematics

1 answer:

Paul [167]3 years ago

5 0

Answer:

b)0.1

pls refer to the pic

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Find the area of the following shape, given its curves are made from parts of circles with diameters indicated on the diagram.

Alexxandr [17]

Answer:

222.51 (when rounded)

Step-by-step explanation:

Pi x r squared is the formula to find the area of a circle

153.94 + 12.57 + 56

= 222.51

You basically find the area of each circle by find their diameters. So half of the length shown on each circle.

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2 years ago

What is the greatest common factor of the numerator and denominator in the expression

adelina 88 [10]

Answer:

Hi i think that the answer is A

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2 years ago

I need help solving this equation -3=9/(5-2k)/5

Amanda [17]

Solve for k by simplifying both sides of the equation, then isolating the variable.

k= 14/5

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3 years ago

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

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2 years ago

Amrita thinks of a number. She doubles it, adds 9 , divides her answer by 3 and finally subtracts 1 . She obtains the same numbe

maks197457 [2]

Answer:

Amrita's number is 6

Step-by-step explanation:

x Double the unknown number

2x Add 9

2x + 9 Divide the answer by 3

2/3x + 3 Subtract 1

2/3x + 2 Set this equal to the original number

2/3x + 2 = x

-2/3x -2/3x Subtract 2/3x from both sides

2 = 1/3x Multiply both sides by 3

6 = x

5 0

3 years ago