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spin [16.1K]
2 years ago
13

24/60 in simplest form explain

Mathematics
2 answers:
const2013 [10]2 years ago
8 0
2/5
In order to find this question quickly, find the greatest common variable between the two numbers. In this case, the number is 12. Now, divide both numbers by their greatest common variable. 24/12= 2 and 60/12= 5.
nekit [7.7K]2 years ago
5 0

24/60 Frist you have to find the greatest common factor which is 12 then you divide 24 and 60 by 12  and you get 2/5
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Ronch [10]
There's no picture or anything about the shape in the question D: if you're able to explain in the comment I'll help you :D
4 0
2 years ago
. The perimeter of a square is 8y – 12. The area of the square is 100 square units. What is the value of y?
evablogger [386]

Answer:

y = 6.5 units

Step-by-step explanation:

Let L be the length of sides of the square.

Given the following data;

Perimeter of square = 8y - 12

Area of square = 100 square units

To find the value of y;

Area of a square = L²

Substituting into the formula, we have;

100 = L²

L = √100

L = 10 units

Mathematically, the perimeter of a square is given by the formula;

Perimeter of a square = 4L

Substituting into the formula, we have;

8y - 12 = 4*10

8y - 12 = 40

8y = 40 + 12

8y = 52

y = 52/8

y = 6.5 units

6 0
2 years ago
Read 2 more answers
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
3 years ago
Which expression is equivalent to 3√32x^8y^10?
denpristay [2]
<span>
3√(32x^8y^10)
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= 12x^4y^5 </span>√2

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3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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