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valentina_108 [34]
3 years ago
14

Find the area of the shaded region in the figure below, if the radius of the outer circle is 8 and the radius of the inner circl

e is 4. Keep your answer in terms of π.

Mathematics
1 answer:
Inessa05 [86]3 years ago
4 0

Answer:

48 π

Step-by-step explanation:

Area of Circle =πr∧2

Big Circle Area = π(8)∧2 = 64π

Small Circle Area =π(4)∧2= 16π

Shaded Region = 64π - 16π = 48π

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Given that f(x) = 6x + 2 and g(x) =2x+4 over 5 solve for g(f(1))
Lady bird [3.3K]
F(1)=6(1)+2
= 6+2
=8

g(f(1))= 2x/5 +4/5
= 2(8)/5 +4/5
= 16/5 +4/5
= 20/5
=4
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X² - 5x + 6 = 0 (R: 2, 3)
Firlakuza [10]
I'm sending you two screenshots of the explanation I hope this helps.

4 0
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Inez waters her plants every two days. She trims them every 15 days. She did both today. When will she do them both again?
shtirl [24]

The answer is 30 days.

The solution for this is to find the least common multiple. By getting the multiple of both numbers.

Multiples of 2:  2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40…..

Multiples of 15: 15,30,45,60,75……..

The Least Common Multiples of 2 and 15 is 30. So, Inez will do them both again in 30 days.

The least common multiple (LCM) of two numbers is the smallest number that is multiple by the both number.

7 0
3 years ago
If p=8×10^5, 5×10^−2. Find p×q​
finlep [7]
Where is q? What should i do?
4 0
2 years ago
Suppose that 5 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 39 cm.
vesna_86 [32]

Answer

given,

work = 5 J

spring stretch form 30 cm to 39 cm

W = \dfrac{1}{2}kx^2

x = 0.39 - 0.30 = 0.09 m

5 = \dfrac{1}{2}k\times 0.09^2

k = \dfrac{5\times 2}{0.09^2}

k = 1234.568 N/m

a) work when spring is stretched from  32 cm to 34 cm

x₂= 0.34 -0.30 = 0.04 m

x₁ = 0.32 - 0.30 = 0.02

W = \dfrac{1}{2}k(x_2^2-x_1^2)^2

W = \dfrac{1}{2}\times 1234.568 \times (0.04^2-0.02^2)^2

W = 0.741 J

b) F = k x

  25 = 1234.568 × x

     x = 0.0205 m

     x = 2.05 cm

3 0
3 years ago
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