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Elena-2011 [213]
3 years ago
10

Fill in the blanks to solve 48 x 8. 48 x 8 = 40 eights + 8 eights

Mathematics
1 answer:
kozerog [31]3 years ago
4 0

Question is:

Fill in the blanks to solve \frac{48}{x} =8.

Answer:

x = 6

Step-by-step explanation:

Given

\frac{48}{x} =8

Required

Solve

Multiply both sides by x

x * \frac{48}{x} =8*x

48 =8*x

Make x the subject

x = \frac{48}{8}

x = 6

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victus00 [196]
1/3 + 7/15 =

4/5


Therefore the answer is 4/5. So Kenneth did 4/5 of his laundry in total.


Hope this helps!
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3 years ago
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) Travis wrote a $53 check for a new pair of shoes. Then, he withdrew $40 from his bank account. What
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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
The volume V of a cube with sides of length x inches is changing with respect to time t (in seconds). When the sides of the cube
goldfiish [28.3K]

Answer:

dV/dt = 3×10^2 × 0.5 = 150 in^3/sec

the volume of the cube is increasing at 150in^3/sec

Step-by-step explanation:

Volume V = length l^3

V = x^3

Differentiating both sides;

dV/dt = 3x^2 dv/dt

Given;

x = 10 in

dx/dt = 0.5 in/sec

dV/dt = 3×10^2 × 0.5 = 150 in^3/sec

the volume of the cube is increasing at 150in^3/sec

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What is the perimeter of the larger rectangle​
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You need to add to find the perimeter

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