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klio [65]
3 years ago
8

11p-4=6p+1 explain steps please ASAP!!!

Mathematics
1 answer:
KatRina [158]3 years ago
8 0
Attached below is the problem that I worked out. And I checked my answer simply by plugging it in

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Given the formula for the perimeter of a rectangle where l represents the length and w represents the width. 2(l + w) What does
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The fact that there are two length sides and two width sides. 
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Joe earns $4,100 per month and pays $870 each month in rent. Approximately what percent of his money is spent on rent?
Scorpion4ik [409]

if we take 4100 to be the 100%, what is 870 off of it in percentage?


\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 4100&100\\ 870&x \end{array}\implies \cfrac{4100}{870}=\cfrac{100}{x} \\\\\\ 4100x=87000\implies x=\cfrac{87000}{4100}\implies x=\cfrac{870}{41}\implies x\approx 21.22

8 0
3 years ago
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
What are the greatest common factors of 48a^6 and 38a?
professor190 [17]
The greatest common factor is 2a
5 0
3 years ago
Read 2 more answers
The points (9, 9) and (5, 1) fall on a particular line. What is its equation in slope-intercept form?
Ilia_Sergeevich [38]
Your equation would be y= 2x-9
3 0
3 years ago
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