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Anni [7]
2 years ago
11

Directions: Calculate the area of a circle using 3.14x the radius

Mathematics
1 answer:
Leokris [45]2 years ago
8 0

\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

\qquad \sf  \dashrightarrow \:r = 2.2 \: mm

Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {4.84}^{}

\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 68.89

\qquad \sf  \dashrightarrow \:area \approx216.31 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 5.8 \div 2

\qquad \sf  \dashrightarrow \:r = 2.9 \:  \: yd

now, let's calculate area ~

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.9)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  8.41

\qquad \sf  \dashrightarrow \:area  \approx26.41 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 1 \div 2

\qquad \sf  \dashrightarrow \:r = 0.5 \:  \: yd

Now, let's calculate area ~

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2}

\qquad \sf  \dashrightarrow \:3.14  \times 0.25

\qquad \sf  \dashrightarrow \:area \approx0.785 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(8)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 64

\qquad \sf  \dashrightarrow \:area = 200.96 \:  \: yd {}^{2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

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Vlad1618 [11]

Answer:

2

Step-by-step explanation:

So first, you know that 4 dozen doughnuts = 48 doughnuts. Next, to find how much you could get for one dollar you have the ratio $24:48 doughnuts and you can divide both sides of this ratio by 24 to get $1:2 doughnuts.

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What is the solution to this system of linear equations 3x 2y 14 5x y 32
Leya [2.2K]

After simplification, the solution of this system of linear equation is (3.85, 12.75).

In the given question we have to find the solution to this system of linear equations 3x-2y=14 or 5x+y=32.

The given system of linear equations is:

3x-2y=14.......................(1)

5x+y=32.......................(2)

Multiply equation 2 by 2, we get

10x+2y=64.......................(3)

Add equation 3 by 1, we get

10x+2y+3x-2y=64-14

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Divide by 13 on both side, we get

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Now put the value of x in equation 2

5(3.85)+y=32

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Subtract 19.25 on both side, we get

y=12.75

So, the solution of this system of equation is (3.85, 12.75).

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Verify that the points are the vertices of a parallelogram, and find its area. A(1, 1, 3), B(2, −5, 6), C(4, −2, −1), D(3, 4, −4
almond37 [142]

Answer:

Area = 71.3 sq unit

Step-by-step explanation:

a) We will use properties of parallelogram to verify the given vertices.

Property: Have a pair of parallel opposite sides

Hence, vectors AB AD BD BC CD AC. A pair should be parallel:

vectors

AB = A - B = (1, 1 , 3) - ( 2 , -5 , 6 ) = (-1 , 6 , -3)

AD = A - D = (1, 1 , 3) - (3, 4 , -4) = (-2 , -3 , 7)

BC = B - C = (2 , -5 , 6) - (4, -2 , -1) = (-2 ,-3 , 7)

CD = C - D = (4, -2 , -1) - (3, 4 , -4) = (1 , -6 , 3)

Hence we can see  that AB //CD and AD // BC as their unit vector co-efficents are identical/scalar multiple of each other.

b)

Equation of line AB = (1,1,3) + t*(-1 , 6 , -3 )

Point E = (1-t , 1+6t , 3 -3t) ... denotes an position of point E on line AB.

Choose a point on line CD, lets suppose C = ( 4 , -2 , -1 )

Vector EC = ( 1-t , 1+6t , 3-3t ) - (4 , -2 , -1 ) = (-3 -t , 3 +6t , 4 -3t)

For EC to be perpendicular to AB then dot product of AB . EC = 0

Hence,

EC . AB = -1 * (-3-t) -2*(3+6t) -3*(4-3t) = 0

3 + t -6 -12t -12+9t = 0

-2t-3 = 0

t = -3/2

Vector EC = (-3 -t , 3 +6t , 4 -3t) = (-1.5 , -6 , 8.5)

Area = magnitude (AB) * magnitude (EC)

Area = sqrt ((-1)^2 + 6^2 + (-3)^2) * sqrt ((-1.5)^2 + 6^2 + (8.5)^2)

Area = sqrt (46) * sqrt (442) / 2

Area = 71.3 unit^2

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Answer:

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Step-by-step explanation:

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Hope it helps!

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