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givi [52]
2 years ago
5

Stan's, Marks and Wayne's ages are consecutive whole numbers. Stan is the youngest, and Wayne is the oldest. The sum of their ag

es is 102. Find their ages
Mathematics
1 answer:
Lisa [10]2 years ago
4 0

Answer: Wayne is 35, Mark is 34, and Stan is 33.

33+34+35 = 102

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Please answer fast The vertices of a rectangle are located at the coordinates (1, 4), (1, 5), (6, 5),
hammer [34]

Answer:

See below ~

Step-by-step explanation:

<u>Drawing the rectangle</u> (Refer attachment)

<u>Horizontal sides</u>

  • There are two x-values present : 1 and 6
  • Find the difference
  • 6 - 1 = 5
  • The horizontal sides of the rectangle are <u>5</u> units long

<u>Vertical sides</u>

  • Two y-values are present : 4 and 5
  • Find the difference
  • 5 - 4 = 1
  • The vertical sides of the rectangle are <u>1</u> unit long

<u>Perimeter</u>

  • 2(Horizontal side + Vertical side)
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2 years ago
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
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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/∝

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Answer:

A and D

Step-by-step explanation:

4×3=12

4×d=4d

12+4d

Then;(4×d)+(3×4)=4d +12

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