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Fittoniya [83]
3 years ago
11

What is the approximate are of the shaded region's?

Mathematics
1 answer:
bixtya [17]3 years ago
4 0

Calculate the area of the squares and the area of the circles, then subtract the area of the circle from the square.

Square with 14 m side:

Area of square = 14 x 14 = 196 square m.

Area of circle inside: 3.14 x 7^2 = 3.14 x 49 = 153.86 square m.

Area of shaded region: 196 - 153.86 = 42.14 square m.

Square with 8 m side:

Area of square: 8 x 8 = 64 square m.

Area of circle: 3.14 x 4^2 = 3.14 x 16 = 50.24 square m.

Area of shaded region: 64 - 50.24 = 13.76 square m.

Round answers as needed.

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Slice for 5x+c=k for x
Degger [83]

Answer:

A - x = (k-c)/5

Step-by-step explanation:

To solve this problem, we would solve for x like we would in 2x = 6 :

5x + c = k

5x = k-c (subtract c from both sides)

x = k/5 - c/5 (divide 5 on both sides)

x = (k-c)/5 (simplify)

:)

4 0
2 years ago
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If f(x)=|3x-4|+2 find f(-10)
Nat2105 [25]

Answer:

f(-10)=36

Step-by-step explanation:

f(x)=|3x-4|+2\\f(-10)=|3(-10)-4|+2\\f(-10)=|-30-4|+2\\f(-10)=|-34|+2\\f(-10)=34+2\\f(-10)=36

Rgds!

6 0
3 years ago
Please help with number 25
Kazeer [188]

Answer:

I think its Damon has 64 stamps or  2^6 (2 to the power of 6)

Step-by-step explanation:

6 0
3 years ago
David and Amrita had an equal number of marbles. After Amrita gave 50 marbles to David, he had 5 times as many marbles as her. F
ZanzabumX [31]

Answer:

Step-by-step explanation:

a = d before Anna gives away 50 marbles.

 

5 (a-50) = a +50 after Anna gives away 50 marbles.

 

5a - 250 = a + 50

 

4a = 300

 

a = 75

 

Anna has 75 marbles at the beginning and so did David.

 

Together they have 150 marbles.

:)

7 0
3 years ago
Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
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