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4 years ago

In a circle, a 90° sector has an area of 36π in2. What is the radius of this circle?

Mathematics

2 answers:

34kurt4 years ago

6 0

Answer:

r = 12 in

Step-by-step explanation:

$A = \frac{1}{4} r^{2} \pi \\\\r = \sqrt{\frac{4A}{\pi } } = \sqrt{\frac{4 * 36\pi }{\pi } } = 2 * 6 = 12\ in$

mamaluj [8]4 years ago

4 0

Answer:

r = 12 in

Step-by-step explanation:

recall that a full circle is represented by a sector that is 360°

in our case, the given sector is 90°

Hence our sector is 90° / 360° = 1/4 of a circle

It is also given that our 90° sector (which we now know is 1/4 of a circle) has an area of 36π in² and recall that the area of a full circle is πr² (where r is the radius).

Hence,

πr² = 4 x area of 90° sector

πr² = 4 x 36π

πr² = 144π (divide both sides by π)

r² = 144

r = √144

r = 12 in

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Sue read 2/3 of the newest mystery book by her favorite author. Jack read 2/6 of the same book. Did Sue and Jack read the same a

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Jack read 2/6 or 1/3. Sue read 2/3, twice as much.

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3 years ago

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One of the roots of the equation 3x^2+7x−q=0 is −5. Find the other root and q.

statuscvo [17]

Answer: The answer is $\textup{The other root is }\dfrac{8}{3}~\textup{and}q=40.Step-by-step explanation: The given quadratic equation is[tex]3x^2+7x-q=0\\\\\Rightarrow x^2-\dfrac{7}{3}x-\dfrac{q}{3}=0.$

Also given that -5 is one of the roots, we are to find the other root and the value of 'q'.

Let the other root of the equation be 'p'. So, we have

$p-5=-\dfrac{7}{3}\\\\\\\Rightarrow p=5-\dfrac{7}{3}\\\\\\\Rightarrow p=\dfrac{8}{3},$

and

$p\times(-5)=-\dfrac{q}{3}\\\\\\\Rightarrow \dfrac{8}{3}\times 5=\dfrac{q}{3}\\\\\\\Rightarrow q=40.$

Thus, the other root is $\dfrac{8}{3}$ and the value of 'q' is 40.

3 0

3 years ago

Two angles of a triangle have the same measure and the third one is 18 degrees greater than the measure of each of the other two

kvv77 [185]

Answer:

72°

Step-by-step explanation:

Let the 2 equal angles be x then the third angle is x + 18

The sum of the 3 angles in a triangle = 180°, thus

x + x + x + 18 = 180, that is

3x + 18 = 180 ( subtract 18 from both sides )

3x = 162 ( divide both sides by 3 )

x = 54

Hence

the largest angle = x + 18 = 54 + 18 = 72°

4 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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3 years ago

a buret is a tool designed to transfer precise amounts of liquid .A buret initially contains 70.00 millimeters of a solution and

katrin [286]

Let:

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Vbk= Volume of the beaker

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x = Number of drips

a = volume of each drip

$\begin{gathered} Vbu=70-ax \\ Vbk=20+ax \\ \text{where:} \\ a=0.05 \\ Vbu=70-0.05x \\ Vbk=20+0.05x \end{gathered}$

after how many drips will the volume of the solution in the buret and beaker be equal ? Vbu = Vbk:

$\begin{gathered} Vbu=Vbk \\ 70-0.05x=20+0.05x \\ \text{Solve for x:} \\ 0.1x=70-20 \\ 0.1x=50 \\ x=\frac{50}{0.1} \\ x=500 \end{gathered}$

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1 year ago