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3 years ago

How to find the area of each of the shaded sectors and round to the nearest hundredth

Mathematics

1 answer:

Mazyrski [523]3 years ago

8 0

Answer:

1-6 you need to write the JHI DEF where the radius center point is in the middle of each shape next to 1-6 some have 2 shapes so use 6 letters there.

Step-by-step explanation:

1. Area = 72 of 360 = 20% of 716.3 = 143.26inch*2 =JHI

2.Area = 19 of 360 = 68.4% of 69.4 = 47.46km^2 =DEF

3. Area = 92 of 360 = 25.56% of 153.9 = 39.33cm^2

4. Area = 226 of 360 = 64.57% of 706.9 = 456.45ft^2

5. Area = 74 of 360 = 20.55% of 153.9 =31.63 inch^2

6. Area = 326 of 360 = 90.56% of 1452 -137.07 = 1314.93m^2

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7 months ago

The length of each leg of an isosceles right triangle is 4 cm. What is the length of the hypotenuse

miss Akunina [59]

<u>Answer:</u>

5.65 cm

<u>Step-by-step explanation:</u>

We are given that the length of each leg of an isosceles right triangle is 4 cm and we are to find the length of the hypotenuse.

For this, we will use the Pythagoras Theorem:

$a ^ 2 = b ^ 2 + c ^ 2$

where $a$ is the hypotenuse.

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3 years ago

Read 2 more answers

(with steps please) Find the inverse Laplace transform, f(t), of the function: 16/(s-4)^3

Stells [14]

Answer: The required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

Step-by-step explanation: We are given to find the inverse Laplace transform, f(t), of the following function :

$F(s)=\dfrac{16}{(s-4)^3}.$

We have the following Laplace formula :

$L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\\\\\\\Rightarrow L^{-1}\{\dfrac{1}{(s-a)^{n+1}}\}=\dfrac{t^ne^{at}}{n!}.$

Therefore, we get

$f(t)\\\\=L^{-1}\{\dfrac{16}{(s-4)^3}\}\\\\\\=16\times\dfrac{t^{3-1e^{4t}}}{(3-1)!}\\\\\\=\dfrac{16}{2}t^2e^{4t}\\\\\\=8t^2e^{4t}.$

Thus, the required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

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3 years ago

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GrogVix [38]

Answer:

Step-by-step explanation:

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