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lisabon 2012 [21]
2 years ago
13

Help I can’t get this ! Please help

Mathematics
1 answer:
Zielflug [23.3K]2 years ago
8 0

Answer:

\sqrt{2^5}

Step-by-step explanation:

First, deal with the product of the two powers of 2 inside the parentheses.

The two factors are powers of 2, so add the exponents.

(2^{\frac{1}{2}} \cdot 2^{\frac{3}{4}})^2 =

= (2^{\frac{1}{2} + \frac{3}{4}})^2

You need a common denominator, 4, to add the fractions.

= (2^{\frac{2}{4} + \frac{3}{4}})^2

= (2^{\frac{5}{4}})^2

Now you have an exponent raised to an exponent. Multiply the exponents and reduce the fraction.

= 2^{\frac{5}{4} \times 2}

= 2^{\frac{10}{4}}

= 2^{\frac{5}{2}}

When a fraction is an exponent, the numerator is an exponent and the denominator is the index of the root. A denominator of 2 means a root index of 2 which means square root.

= \sqrt{2^5}

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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:

\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

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