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Wittaler [7]
3 years ago
15

420 miles In 6 hours

Mathematics
1 answer:
Irina18 [472]3 years ago
6 0

Answer:

yeah me too

Step-by-step explanation:

6+36

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WHOEVER ANSWERS THIS WILL GET BRAINLIEST!!! SERIOUS ANSWERS ONLY!!!
ale4655 [162]

Answer:

496.8926

Step-by-step explanation:

20 = 100(1/2)^(t/214)

Divide both sides by 100

20/100 = 100(1/2)^(t/214) / 100

1/5 = (1/2)^(t/214)

log((1/2)^(t/214) = log(1/5)

(t/214) x log(1/2) = log (1/5)

(t/214) = log (1/5)

           ÷ log (1/2)

t/214 = 2.321928

(214) x (t/214) = (214) x (2.321928)

t = 496.892592

I hope this helps!

6 0
2 years ago
12.5*<img src="https://tex.z-dn.net/?f=12.5%20%2A17.9%3D" id="TexFormula1" title="12.5 *17.9=" alt="12.5 *17.9=" align="absmiddl
kkurt [141]
The answer is 223.75 because you multiply 12.5 and 17.9
5 0
3 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
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
The slope of a line is –2 and its y-intercept is (0, 3). What is the equation of the line that is parallel to the first line and
notsponge [240]

Answer:

D. y=-2x-6

Step-by-step explanation:

<u><em>First start with what we know....</em></u>

y = -2x + 3 (Slope Intercept Form)

<u><em>Because of this we can eliminate B.  </em></u>

<u><em>Parallel means that the lines wouldn't be touching which means they should have the same slope and the only one with the same slope is D. </em></u>

7 0
3 years ago
Read 2 more answers
Such that |x-a|&lt;\varepsilon and $x-b|&lt;\varepsilon. can $a$ be countable? can $a$ be uncountable?
Brrunno [24]
It will be uncountable

3 0
3 years ago
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