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ipn [44]
3 years ago
10

The circumference of a circle is 6π inches. What is the area of the circle? Alternative Text A. 3π in.2 B. 9π in.2 C. 12π in.2 D

. 36π in.2
Mathematics
1 answer:
alukav5142 [94]3 years ago
8 0

Answer:

B) 9π in^2

Step-by-step explanation:

Circumference of a circle is C = πD

Divide 6π by π to get 6 then divide by 2 to get the radius of the circle

Area of a circle is A = πr^2

6/2 = 3 3 = r

π(3)^2 = 9π

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I need the answer please
Veseljchak [2.6K]
The trick to solving this problem is to know and remember that the sum of all the interior angles of a triangle is always 180 degrees.

Thus, (6x+1) + (5x-17) + (9x-24) = 180.

20x = -40, so x = -2 (answer)
7 0
3 years ago
Which of the following statements is true?
Inessa05 [86]

Answer: the number -5 is a real number

Real numbers are both negative and positive numbers. -5 is a negative number. Since real numbers include negative numbers, -5 is the answer.

Hope this helps!

7 0
3 years ago
What is the kinetic Energy of a camion of 5,000 kg and turn at 100km/h ?
Alex787 [66]

Step-by-step explanation:

Given

mass (m) = 5000 kg

velocity (v) = 100 km/h = 100* 1000/ 3600 = 27.78m/s

Now

Kinetic energy

= 1/2 mv²

= 1/2 * 5000 * 27.78²

= 1929321 joule

6 0
3 years ago
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

============================================================

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
Which expressions are equivalent to –
olga_2 [115]

Answer:

The answer to your question is  10¹/⁴

Step-by-step explanation:

Expression

                            \frac{10}{10^{3/4}}

To solve this problem use the law of exponents.

-Use the division rule of exponents.

If the base if the same in the numerator and denominator, it stays the same and the exponents are subtracted.

                            10¹ ⁻ ³/⁴ = 10¹/⁴   This is the result

7 0
3 years ago
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