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3 years ago

6

The area of a circle is 50.24 square centimeters what is the diameter

2 answers:

quester [9]3 years ago

6 0

Answer:

it is 8 feet

Step-by-step explanation:

stiv31 [10]3 years ago

4 0

I hope this helps you

Area=pi×r^2

50.24=3.14×r^2

r^2=16

r=4

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A. Find y in terms of x if

givi [52]

Answer:

A. $y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

B.Therefore the solution is defined on the interval:

6≤ x ≤ +∞

Step-by-step explanation:

Given differential equation is

$\frac{dy}{dx} = x^4y^{-1}$

$\Rightarrow \frac{dy}{dx} =\frac{ x^4}{y}$

$\Rightarrow y dy = x^4dx$

Integrating both sides:

$\Rightarrow\int y dy = \int x^4dx$

$\Rightarrow \frac{y^2}{2} =\frac{x^5}{5} +C$ ........(1)

Given y(0)= 6

it means y=0 when x=6

Putting x=6 and y=0 in the equation (1)

$\therefore \frac{0^2}{2} =\frac{6^5}{5} +C$

$\Rightarrow C=-\frac{6^5}{5}$

Equation (1) become:

$\Rightarrow \frac{y^2}{2} =(\frac{x^5}{5} -\frac{6^5}{5})$

$\Rightarrow y} =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

Therefore $y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

(B)

The quantity under root must greater than or equal to zero.

Therefore,

$2(\frac{x^5}{5} -\frac{6^5}{5})} \geq0$

$\Rightarrow \frac{x^5}{5} -\frac{6^5}{5}} \geq0$

$\Rightarrow \frac{x^5}{5} \geq\frac{6^5}{5}}$

$\Rightarrow x^5\geq {6^5$

$\Rightarrow \sqrt[5]{x^5} \geq \sqrt[5]{6^5}$

$\Rightarrow x\geq {6$

When x≥0 then the value of $2(\frac{x^5}{5} -\frac{6^5}{5})}$ is real other wise the value of $2(\frac{x^5}{5} -\frac{6^5}{5})}$ is imaginary.

Therefore the solution is defined on the interval:

6≤ x ≤ +∞

8 0

4 years ago

Please answer this correctly without making mistakes

Ber [7]

66.7

you will get the answer

5 0

3 years ago

Read 2 more answers

What value of m satisfies the equation<br> m-2=2-m?<br>

zhuklara [117]

Hello,

Answer: m = 2

Step-by-step explanation:

m - 2 = 2 - m

⇔ m - 2 + m = 2 - m + m

⇔ 2m - 2 = 2

⇔ 2m - 2 + 2 = 2 + 2

⇔ 2m = 4

⇔ 2m/2 = 4/2

⇔ m = 2

7 0

2 years ago

Read 2 more answers

There were several sunny days this year before my birthday. Since my birthday, there have been 171717 more sunny days. My birthd

Lesechka [4]

Answer:

66 sunny days before birthday

Step-by-step explanation:

Total sunny days after birthday = 17

Total sunny days in year = 83

Birthday was not sunny

Total sunny days before birthday = 83 -17

= 66 sunny days

Read more on Brainly.in - https://brainly.in/question/6937259#readmore

4 0

4 years ago

The Environmental Protection Agency records data on the fuel economy of many different makes of cars. Data on the mileage of 20

artcher [175]

Answer:

A) 8.5 miles per gallon

Step-by-step explanation:

Given

$x = \{12\ 13\ 15\ 16\ 16\ 17\ 18\ 18\ 19\ 19\ 20\ 20\ 22\ 23\ 24\ 26\ 26\ 27\ 27\ 29\}$

$n = 20$

Required

The IQR

This is calculated as:

$IQR = Q_3 - Q_1$

Calculate $Q_1$

$Q_1 = \frac{1}{4}(n + 1)th$

$Q_1 = \frac{1}{4}(20 + 1)th$

$Q_1 = \frac{1}{4}(21)th$

Remove bracket

$Q_1 = 5.25th$

This means that:

$Q_1$ is the mean of 5th and 6th item

From the given data:

$5th \to 16$

$6th \to 17$

So, we have:

$Q_1 = \frac{1}{2}(16 + 17)$

$Q_1 = \frac{1}{2}(33)$

$Q_1 = 16.5$

$Q_1 = 16.25$

Calculate $Q_3$

$Q_3 = \frac{3}{4}(n + 1)th$

$Q_3 = \frac{3}{4}(20 + 1)th$

$Q_3 = \frac{3}{4}(21)th$

Remove bracket

$Q_3 = 15.75th$

This means that:

$Q_3$ is the mean of 15th and 16th item

From the data:

$15th \to 24$

$16th \to 26$

So:

$Q_3 = \frac{1}{2}(24 + 26)$

$Q_3 = \frac{1}{2}(50)$

$Q_3 = 25$

Recall that:

$IQR = Q_3 - Q_1$

$IQR = 25 -16.5$

$IQR = 8.5$

7 0

3 years ago