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2 years ago

The radius of a circle is 3 inches . What is the circumference of the circle ?

Mathematics

1 answer:

Vlada [557]2 years ago

3 0

Answer:

Option C

Step-by-step explanation:

Circumference of a circle is:

$C=2\pi r$

r - radius

We are given the radius of 3 inches.

$C=2\pi r\\\\C=2(3)\pi \\\\C=\boxed{6\pi }$

Option C should be the correct answer.

Hope this helps!

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Tim installs 50 square feet of his floor in 45 minutes. At this rate, how long does it take him to install 495 square feet?

77julia77 [94]

Answer:

99.5 minutes

Step-by-step explanation:

50=45

495=?

495×45÷50=99.5

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If you double a number and then add 32, the result is 80. what is the number?

maw [93]

80-32 ÷ 2= 24
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2 years ago

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Whats the product of (3-2i) and (7+6i)?

Alona [7]

Answer: the product is = 33+4i

Step-by-step explanation:

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3 years ago

The terminal point of Dis ( 1/2,3/2). What is cos 0?

nydimaria [60]

Answer:

(c). $\frac{1}{2}$

Step-by-step explanation:

The cosine is the "x" coordinate of the terminal point on the unit circle.

Thus the answer is (c).  $\frac{1}{2}$

5 0

3 years ago

Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),

amm1812

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

Evaluating the iterate integral $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.$

Thus, $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=24$

7 0

2 years ago