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Harrizon [31]
2 years ago
5

Y=x^2+8 Find the Domain, explain your answer

Mathematics
1 answer:
alekssr [168]2 years ago
3 0

Step-by-step explanation:

look at the attachment above ☝️

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What is 2.26 (6 repeating) as a fraction?
Gekata [30.6K]

Answer:

34/15

Step-by-step explanation:

I recognized the form of this decimal as a number divided by 3. However, when I multiplied it by 3, I got 6.8. This means that the fraction was not fully simplified. Multiplying by 5, I got 34. Since I had to multiply by 3 and then by 5 to get there, I multiplied by a total of 15. Hope this helps!

3 0
3 years ago
PLEASE HELP!! i’m literally desperate lol. answers are 18.1, 18.7, 21.6, or 19.3
romanna [79]

I think the answer is 18.7

6 0
3 years ago
brenda savings account principal 850, 5.5% annual interest rate and is compounded quarterly for 1 year. what is the compound int
Sedbober [7]
A = $ 861.69

Equation:
A = P(1 + rt)
Calculation:
First, converting R percent to r a decimal
r = R/100 = 5.5%/100 = 0.055 per year,
putting time into years for simplicity,
1 quarters ÷ 4 quarters/year = 0.25 years,
then, solving our equation

A = 850(1 + (0.055 × 0.25)) = 861.6875 
A = $ 861.69

The total amount accrued, principal plus interest,
from simple interest on a principal of $ 850.00
at a rate of 5.5% per year
for 0.25 years (1 quarters) is $ 861.69.
3 0
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
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 of the following relations is NOT a function?
Alexxandr [17]

Answer:

It is A.

Step-by-step explanation:

In function A we see that there are 2 ordered pairs with value 2 in the first position with y values 6 and 2. So this is not a function.

5 0
3 years ago
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