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3 years ago

Why dint you look at that another question

Mathematics

1 answer:

7nadin3 [17]3 years ago

5 0

Answer:

B. 4

Step-by-step explanation:

36/9=4, since 9*4=36. Therefore, the correct answer is B, 4. Hope this helps!

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The perimeter of a shape is 25.71 cm calculate the value of the radius X takes pie to be 3.142

Mariulka [41]

Answer:

5 cm

Step-by-step explanation:

Given a semicircle with perimeter 25.71 cm, we are required to find the radius of the circle.

Circumference of a Semicircle

=Length of the Circular part+Length of the diameter

$=\pi r+2r$

$If \: \pi=3.142, C=25.71\\Then:\\25.71=3.142r+2r\\r(3.142+2)=25.71\\5.142r=25.71\\r=25.71 \div5.142\\r=5cm$

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2 years ago

Plsss helppp its worth 20 points

Mazyrski [523]

Answer:

a. sides BC & EF

b. <A & <D

2a. A / B, not that sure

8 0

2 years ago

How to write 325809 in expanded form

BartSMP [9]

Three hundred twenty five thousand eight hundred nine

7 0

3 years ago

Read 2 more answers

Given the following formula, solve for l.

scoray [572]

P=2(l + b)
P=2l + 2b
P-2b=2l
P-2b/2=l
C is your answer

8 0

2 years ago

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Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

3 years ago