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3 years ago

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A circle has a sector with area \dfrac{64}{5}\pi 5 64 πstart fraction, 64, divided by, 5, end fraction, pi and central angle o

f \purple{\dfrac{8}{5}\pi} 5 8 πstart color #9d38bd, start fraction, 8, divided by, 5, end fraction, pi, end color #9d38bd radians . What is the area of the circle?

1 answer:

Marina86 [1]3 years ago

8 0

Answer:

Area of circle is $\dfrac{1024\:\pi}{25}$

Step-by-step explanation:

Since angle $\theta$ is given in radians, so using formula for area of sector in radians is given as,

$Area\:of\:sector=\dfrac{1}{2}\:r^{2}\:\theta$

Given area of sector and angle. Therefore substituting the values in above formula,

$\dfrac{64\pi}{5}=\dfrac{1}{2}\:r^{2}\:\left (\dfrac{8\pi}{5}\right )$

Cancelling out the common term,

$16=\dfrac{1}{2}\:r^{2}$

Multiplying with 2 on both sides,

$16=r^{2}$

Therefore value of $r^{2}$ is $16$

Formula for area of circle is,

$Area\:of\:circle=\pi\:r^{2}$

Substituting the value,

$Area\:of\:circle=\pi\times16$

$Area\:of\:circle=16\pi$

Therefore area of circle is $16\pi$

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babymother [125]

Answer: 3

Step-by-step explanation:

1. 2²=4

2. 4*5=20

3. 180/20=9

4. √9=3

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4 0

3 years ago

Read 2 more answers

Given teo circles whose radii are 9" and 24", the ratio of thuer circumfrence is?

saul85 [17]

Answer:

The ratio is 3:8

Step-by-step explanation:

The circumference of a circle can be calculated using the formula;

C = 2 * pi * r

for 9 inches;

C = 2 * pi * 9 = 18 pi inches

For 24 inches;

C = 2 * pi * 24 = 24 pi inches

The ratio is thus;

18 pi : 48 pi

divide through by 6 pi

= 3: 8

5 0

3 years ago

Simplify: 18q2−45q+25/9q2−25.

Svet_ta [14]

Answer:

$\frac{(6q-5)}{(3q+5)}$

Step-by-step explanation:

Given: $\frac{18q^{2}-45q+25 }{9q^{2}-25 }$

Factorizing the numerator and the denominator, we have:

18 $q^{2}$ - 45q + 25 = 18 $q^{2}$ - 30q - 15q + 25

= (3q - 5)(6q - 5)

and

9 $q^{2}$ - 25 = (3q - 5)(3q + 5)

So that,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}$

= $\frac{(6q-5)}{(3q+5)}$

Therefore,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(6q-5)}{(3q+5)}$

5 0

3 years ago

Ella has a part-time job selling cupcakes at a cart in the park. She is paid $25 for a shift plus a fee of $0.25 for each cupcak

mojhsa [17]

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Answer:

A) t = 0.25c +25

B) c = 52

C) 3 weekday shifts

Step-by-step explanation:

A. The amount Ella earns is $0.25 per cupcake, so for c cupcakes, that is ...

0.25c

In addition, she is paid $25, so her total amount she is paid is ...

t = 0.25c +25

__

B. In order to earn $38, the number of cupcakes sold must be ...

38 = 0.25c +25

13 = 0.25c . . . . . . subtract 25

52 = c . . . . . . . . . multiply by 4

Ella sold 52 cupcakes to earn $38.

__

C. On one of the average weekdays, Ella could be expected to earn ...

t = 0.25(36) +25 = 34

Then for 3 weekday shifts, her earnings would be 3($34) = $102. (It would take Ella 3 weeks to earn $300.)

On one of the average weekend shifts, Ella could be expected to earn ...

t = 0.25(62) +25 = 40.50

Then for 2 weekend shifts, her earnings would be 2($40.50) = $81. (It would take Ella 4 weekends to earn $300.)

Ella would earn $300 more quickly by working weekday shifts.

8 0

3 years ago

Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1

Shkiper50 [21]

Answer:

$\displaystyle -\frac{1}{2} \leq x < 1$

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

$\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0$

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

$\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0$

Simplifying by x-1 and taking x=1 out of the possible solutions:

$\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0$

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

$(6x^3+14x^2+10x+12)$

is always positive and doesn't affect the result. It can be neglected. The expression

$(x-\frac{1}{2})^2$

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

$\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0$

For the expression to be negative, both signs must be opposite, that is

$(x+\frac{1}{2})\geq 0, (x-1)$

Or

$(x+\frac{1}{2})\leq 0, (x-1)>0$

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

$\displaystyle -\frac{1}{2} \leq x < 1$

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

$\boxed{\displaystyle -\frac{1}{2} \leq x < 1 }$

7 0

3 years ago