Please help me on 5 I don’t get it LOL 11 point pleaseeee

Triangle BCD is equivalent AG=1.find the perimeter of BCD

Ymorist [56]

AG=1 => BG=3 => CG=sqrt3 => CD = 2sqrt3 => perimeter = 3*CD= 6sqrt3

4 0

4 years ago

A study conducted on 40,000 words in the English language found that about 6,670 words started with the letter T, which is the m

melisa1 [442]

To answer this problem, first let us calculate for the theoretical probability obtained from the study conducted:

Theoretical Probability = 6,670 / 40,000

Theoretical Probability = 0.16675 = 0.17

From the given choices, the list of words which can give this similar amount of probability is:

“to, sarcastic, epithet, for, less, corruptible, than, the, offspring, open, any, primary, loyalty, table, out, whom, example, who, mention, say”

In this set of words, the number of words starting with letter T is 4. Calculating for the probability of getting a word starting with T in this set is:

Sample probability = 4 / 20

<span>Sample probability = 0.20</span>

5 0

3 years ago

Read 2 more answers

The sum of two numbers is 8, and their product is 15. What is the largest number?

AfilCa [17]

Hello there!

The sum of two numbers (x + y) is 8 (=8).
Their product (x * y) is 15 (=15).
x + y = 8
x * y = 15

First, we know that the two number must be positive and under 8, as the product is positive.

Let's list out some numbers that are from 1 to 8 that sum up to 8;
1 + 7
2 + 6
3 + 5
4 + 4

Now, we need to multiply the two numbers that add up to 8, and see which makes a product of 15.
1 * 7 = 7, this does not work.
2 * 6 = 12, this does not work.
3 * 5 = 15, this works!
4 * 4 = 16, this does not work.

Since we are looking for the larger number in our pairs, let's pick the largest one out of the correct pair, which is 5.

The largest number is 5.

I hope this helps!

3 0

3 years ago

If 12 1/2% of a sum of money is $40, what is the total sum of money?

GarryVolchara [31]

Answer:

i think its 320

Step-by-step explanation:

i looked it up

8 0

3 years ago

Read 2 more answers

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

2 years ago