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Colt1911 [192]
3 years ago
12

PLZ HELP I'VE BEEN STUCK FOR A LONG TIME PLZ HELP

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

c/d = 6

Step-by-step explanation:

Use the second equation.

c - 6d = 0

Add 6d to both sides.

c = 6d

Divide both sides by d.

c/d = 6

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I need help finding the values of the last boxes shown in the image.
Bingel [31]

The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

<h3>What is the volume of the solid revolution on the X-axis?</h3>

The volume of a solid is the degree of space occupied by a solid object. If the axis of revolution is the planar region's border and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

In the graph, the given straight line passes through two points (0,0) and (2,8).

Therefore, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

where:

  • (x₁, y₁) and (x₂, y₂) are two points on the straight line

Thus, from the graph let assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8), we have:

\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}

y = 4x

Now, our region bounded by the three lines are:

  • y = 0
  • x = 2
  • y = 4x

Similarly, the change in polar coordinates is:

  • x = rcosθ,
  • y = rsinθ

where;

  • x² + y² = r²  and dA = rdrdθ

Now

  • rsinθ = 0   i.e.  r = 0 or θ = 0
  • rcosθ = 2 i.e.   r  = 2/cosθ
  • rsinθ = 4(rcosθ)  ⇒ tan θ = 4;  θ = tan⁻¹ (4)

  • ⇒ r = 0   to   r = 2/cosθ
  •    θ = 0  to    θ = tan⁻¹ (4)

Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

Learn more about the determining the volume of solids bounded by region R here:

brainly.com/question/14393123

#SPJ1

5 0
2 years ago
A cylinder has a radius of 14m and height of 6m. What is the exact volume of the cylinder?
bagirrra123 [75]
3694.51 is the answer because the formula you have to use is V=pi r^2h.
V=pi14^2*6 
V=pi*196*6
V=pi*1176
V=3694.51
3 0
3 years ago
Read 2 more answers
Allyson accidentally missed her turn on her way to visit her aunt. By the time she realized her mistake, she had been driving fo
olya-2409 [2.1K]

Answer:

can u convert this to miles?

Step-by-step explanation:

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Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg
Lady bird [3.3K]

Answer:

1813.137 KJ

Step-by-step explanation:

potential energy of the body = mgH

where m is mass in Kg , g= 9.81 m/sec^2 and H= height in m

here m= 1000 kg, g= 9.81 m/s^2  and H= 607 feet = 607×0.305= 185.135 m

hence the potential energy p= 1000×9.81×185.135= 1813137.2 J

= 1813.137 KJ

hence the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass as 1000 kg is = 1813.137 KJ

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