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Snowcat [4.5K]
2 years ago
7

ASAP HELP! BRAINLIEST! can someone help me with this? i need work shown. will mark brainliest!

Mathematics
1 answer:
ki77a [65]2 years ago
5 0

Answer:

  4x^2 +12x +44 remainder 161x +84

Step-by-step explanation:

At each step, the quotient term is the ratio of the leading dividend term to the leading divisor term. The first quotient term, for example, is ...

  (4x^4)/(x^2) = 4x^2

The quotient term found this way is multiplied by the divisor and subtracted from the dividend. The difference is the new dividend and the process repeats.

You're done when the degree of the dividend is less than the degree of the divisor. This remainder can be expressed as a fraction with the divisor as the denominator.

  \dfrac{4x^4+5x-4}{x^2-3x-2}=4x^2+12x+44+\dfrac{161x+84}{x^2-3x-2}

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solniwko [45]
It would be less because when you multiply something by 1/2 you are basically dividing it by 2, finding half. So when you multiply 9 8/9 by 1/2 your product would be less than 9 8/9
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Need your help now please; the assignment is due tonight and this is the last problem I am having trouble with. The red box is t
Kobotan [32]

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 (R) on the X-axis?</h3>

If the axis of revolution is the boundary of the plane region 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.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

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

here:

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

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, 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

Therefore;

  • 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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