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Lera25 [3.4K]
3 years ago
9

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

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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
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