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1 year ago

What is the lowest common multiple of 14 & 28

Mathematics

1 answer:

siniylev [52]1 year ago

6 0

To find the LCM:

1. List the multiples of each number:

Multiples of 14:

$14\colon14,28,42,56,70$

Multiples of 28:

$28\colon28,56,84,112,140$

2. The LCM is the smallest number that appears in the two lists above.

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5<br> 0 = - in<br> =<br> in quadrant II<br> Given cos =<br> Find sin<br> 3

____ [38]

Answer:

sinΘ = $\frac{2}{3}$

Step-by-step explanation:

using the identity

sin²x + cos²x = 1 ( subtract cos²x from both sides )

sin²x = 1 - cos²x ( take square root of both sides )

sinx = ± $\sqrt{1-cos^2x}$

given

cosΘ = - $\frac{\sqrt{5} }{3}$ , then

sinΘ = ± $\sqrt{1-(-\frac{\sqrt{5} }{3})^2 }$

= ± $\sqrt{1-\frac{5}{9} }$

= ± $\sqrt{\frac{4}{9} }$

= ± $\frac{2}{3}$

since Θ is in quadrant II where sinΘ > 0 , then

sinΘ = $\frac{2}{3}$

7 0

2 years ago

Which of the following expressions has a value that is less than –2?

natali 33 [55]

Answer: A

Step-by-step explanation: 0.25−2.50=-2.25

3 0

3 years ago

For which value(s) of x are the triangles congruent?

Varvara68 [4.7K]

Step-by-step explanation:

The solutions are 2 and 5

Because it's asking what would make the triangles congruent, you would set up the equation like this because the angles (angle 3 and angle 4) need to be equal:

x^2 = 7x - 10

Next, you add 10 to both sides. This is so that you can move it to the other side, addition is the inverse of subtraction.

x^2 + 10 = 7x

Now subtract 7x from both sides. Subtraction is the inverse of addition. You do this to get it on the other side so you can factor it. You can move the 10 and 7x to the other side in any order or at the same time, I just did it like this.

x^2 - 7x + 10 = 0

Now, factor. I don't really know how to explain factoring, you just get a feel for it with a lot of practice.

(x - 2)(x - 5) = 0

You can use FOIL to check this if you want to. x(x) is x^2, -2(-5) is 10), -2x - -5x is -7x. Now, find what you need to do to make what's in each of the groups of parentheses equal to 0.

x - 2 = 0

x = 2

One of the solutions is 2, because you add 2 to x to get 0.

x - 5 = 0

x = 5

The other solution is 5, because you add 5 to x to get 0. Lastly, check your solutions by plugging them in to the original equation.

2^2 = 7(2) - 10

4 = 14 - 10

So 2 is definitely a solution.

5^2 = 7(5) - 10

25 = 35 - 10

5 is also a solution.

Hope that helps :]

8 0

2 years ago

Evaluate the upper and lower sums for f(x) = 1 + x2, −1 ≤ x ≤ 1, with n = 3 and 4.

Elenna [48]

I'm going to assume that your function is f(x) = 1 + x^2 (NOT x2).

I suspect you're trying to estimate the "area under the curve of f(x) = 1 + x^2. You need to use this or a similar description to explain what you're doing.

Also, you need to specify whether you want "left end points" or "right end points" or "midpoints." Again I must assume you want one or the other (and will assume that you meant "left end points").

First, let's address the case n=3. You must graph f(x) = 1 + x^2 between -1 and +1. We will find the "lower sum," using "left end points." The 3 x-values are {-1, -1/3, 1/3}. Evaluate the function f(x) = 1 + x^2 at these 3 x-values. Keep in mind that the interval width is 2/3.
The function (y) values are {0, 2/3, 4/3}.

Sorry, Michael, but I must stop here and await clarification from you regarding what you've been told to do in this problem. Otherwise too much guessing (regarding what you meant) is necessary. Please review the original problem and ensure that you have copied it exactly as presented, and also please verify whether this problem does indeed involve estimating areas under curves between starting and ending x-values.

7 0

3 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.

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

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