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lana66690 [7]
3 years ago
5

190 = 200^b, make b the subject​

Mathematics
1 answer:
MArishka [77]3 years ago
3 0

Answer:

b = \frac{ln190}{ln200}

Step-by-step explanation:

Using the rule of logarithms

log x^{n} ⇔ nlogx

Given

190 = 200^{b} ( take the natural log ln of both sides )

ln190 = ln200^{b} = bln200 ( divide both sides by ln200 )

\frac{ln190}{ln200} = b

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I need help finding the values of the last boxes shown in the image.
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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:

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Now, our region bounded by the three lines are:

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Similarly, the change in polar coordinates is:

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where;

  • x² + y² = r²  and dA = rdrdθ

Now

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  • 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

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