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3 years ago

What is the circumference of five

Mathematics

2 answers:

RideAnS [48]3 years ago

4 0

Answer:

circle

Step-by-step explanation:

Alex3 years ago

4 0

Answer:

if 5 is the radius

(circumference formula is 2*pi*r)

so it'll be 2*pi*5 = 31.41592654...

if 5 is the diameter, then the radius is 2.5 (radius= diameter/2)

so the circumference will be 2*pi*2.5 = 15.70796327

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A study concludes that 15 percent of all college students attend a gym class.

ra1l [238]

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2 years ago

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Evaluate the integral Integral from 0 to 1 Integral from 0 to 3 Integral from 3 y to 9 StartFraction 6 cosine (x squared )Over 5

Natalka [10]

Answer:

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$ $=\frac{18}{5}(1+\frac{sin2}{2})$

Step-by-step explanation:

cosine x²= cos x²

Rule

$\int x^ndx= \frac{x^{n+1}}{n+1}+c$

$\int cos \ mx \ dx = \frac{sin \ mx}{m}+c$
$\int \frac{1}{\sqrt x}dx = \frac{\sqrt x}{\frac{1}{2}} +c= 2\sqrt x+c$

Given that,

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$

$=\int ^1_0[\int^3_0(\int^9_{3y} \frac{6cos x^2}{5\sqrt z}dz)dy]dz$

$=\int^1_0[\int^3_0([\frac{6cos x^2 \times \sqrt z}{5\times \frac{1}{2}}]^9_{3y})dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( \sqrt 9-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( 3-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\frac{12cos x^2 \times( 3y-\frac{\sqrt{3}y^\frac{3}{2}}{\frac{3}{2}})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 3.3-\frac{2\sqrt{3}.3^\frac{3}{2}}{3})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 9-6)}{5}]dx$

$=\frac{18}{5}\int^1_02cos x^2dx$

$=\frac{18}{5}\int^1_0(1+cos2x)dx$

$=\frac{18}{5}[(x+\frac{sin2x}{2})]^1_0$

$=\frac{18}{5}(1+\frac{sin2}{2})$

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3 years ago

2x+y=-4 ×+y=-7 solve the system by substitution

ANEK [815]

Answer:

x = 3, y = -10

(3.-10)

Hopefully this helps!

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3 years ago

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2 years ago

Express log2 64-log2 4 as a simple logarithm

Triss [41]

Express log(2)64-log(2)4 as a single logarithm.
Simplify as possible.
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2 years ago