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hichkok12 [17]
3 years ago
15

Four less than three times a number is greater than two times the number minus one

Mathematics
2 answers:
Andreyy893 years ago
6 0
4-3x>2x-1
Hope this helps!!
kkurt [141]3 years ago
5 0
4-3x> 2x-1
-5x<-5
x<1

Hope this helps
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The value of the surface area of the cylinder is equal to the value of the volume of the cylinder. Find the value of x.
zlopas [31]

Answer:

x = 2.5

Step-by-step explanation:

Surface area of cylinder = 2πr(h + r)

Volume of cylinder = πr²h

Given that S.A = Volume of the cylinder, therefore, we have:

2πr(h + r) = πr²h

Radius (r) is given as 2.5 cm

height (h) = x cm

Input the values and solve for x

2πr(h + r) = πr²h

2πr(h + r) = πr(rh)

2(h + r) = rh (πr cancels πr)

2(x + 2.5) = 2.5*x

2x + 5 = 2.5x

Subtract 2x from both sides

2x + 5 - 2x = 2.5x - 2x

5 = 0.5x

Divide both sides by 0.5

\frac{5}{0.5} = \frac{0.5x}{0.5}

2.5 = x

x = 2.5

8 0
3 years ago
The endpoints of a line segment graphed on the coordinate system are (−7, 3) and (2, 5). What is the range of the graph?
Eduardwww [97]
I think the answer would be C
4 0
3 years ago
If f(x) = 10 + 50x, solve for x when f(x) = 310.
weqwewe [10]

x = 6

given f(x) = 310, we obtain the equation

10 + 50x = 310 ( subtract 10 from both sides )

50x = 300 ( divide both sides by 50 )

x = \frac{300}{50} = 6


5 0
3 years ago
F(x) = square root 2x
dezoksy [38]

\\ \sf\longmapsto (f.g)(x)

\\ \sf\longmapsto f(x).g(x)

\\ \sf\longmapsto \sqrt{2x}.\sqrt{50x}

\\ \sf\longmapsto \sqrt{100x^2}

\\ \sf\longmapsto 10x

5 0
2 years ago
Can someone explain step by step how to do this problem? Thanks! Calculus 2
Ganezh [65]

Answer:

  1.314 MJ

Step-by-step explanation:

As water is removed from the tank, decreasing amounts are raised increasing distances. The total work done is the integral of the work done to raise an incremental volume to the required height.

There are a couple of ways this can be figured. The "easy way" involves prior knowledge of the location of the center of mass of a cone. Effectively, the work required is that necessary to raise the mass from the height of its center to the height of the discharge pipe.

The "hard way" is to write an expression for the work done to raise an incremental volume, then integrate that over the entire volume. Perhaps this is the method expected in a Calculus class.

<h3>Mass of water</h3>

The mass of the water being raised is the product of the volume of the cone and the density of water.

The cone volume is ...

  V = 1/3πr²h . . . . . . for radius 2 m and height 8 m

  V = 1/3π(2 m)²(8 m) = 32π/3 m³

The mass of water in the cone is then ...

  M = density × volume

  M = (1000 kg/m³)(32π/3 m³) ≈ 3.3510×10^4 kg

<h3>Center of mass</h3>

The center of mass of a cone is 1/4 of the distance from the base to the point. In this cone, it is (1/4)(8 m) = 2 m from the base.

<h3>Easy Way</h3>

The discharge pipe is 2 m above the base of the cone, so is 4 m above the center of mass. The work required to lift the mass from its center to a height of 4 m above its center is ...

  W = Fd = (9.8 m/s²)(3.3510×10^4 kg)(4 m) = 1.3136×10^6 J

<h3>Hard Way</h3>

As the water level in the conical tank decreases, the remaining volume occupies a space that is similar to the entire cone. The scale factor is the ratio of water depth to the height of the tank: (y/8). The remaining volume is the total volume multiplied by the cube of the scale factor.

  V(y) = (32π/3)(y/8)³

The differential volume at height y is the derivative of this:

  dV = π/16y²

The work done to raise this volume of water to a height of 10 m is ...

  (9.8 m/s²)(1000 kg/m³)(dV)((10 -y) m) = 612.5π(y²)(10 -y) J

The total work done is the integral over all heights:

  \displaystyle W=612.5\pi\int_0^8{(10y^2-y^3)}\,dy=\left.612.5\pi y^3\left(\dfrac{10}{3}-\dfrac{y}{4}\right)\right|_0^8\\\\W=612.5\pi\dfrac{2048}{3}\approx\boxed{1.3136\times10^6\quad\text{joules}}

It takes about 1.31 MJ of work to empty the tank.

8 0
1 year ago
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