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luda_lava [24]
3 years ago
7

Need help please fast and urgent

Mathematics
1 answer:
Airida [17]3 years ago
6 0
The answer is b bro i belive

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Plot the data for the functions f(x) and g(x) on a grid
wariber [46]
The graph (by some miracle) has been uploaded for you. It is just about the first time I've done this sort of thing, and I've answered nearly 800 questions. 

The first thing you have to do is study the graph. The two functions are
f(x) = 4^x That's the curved graph. (in red)
g(x) = x + 4. That's the straight line. (in blue)

You know that the first one is not a linear relationship because the x values go from integer values -2 to 2 (including 0). The y values are a bit different. They go from 1/16 to 16 with those integer values. So you could try y = 4^(-x). It doesn't work, but you could try it. It gives the table numbers for y in the reverse order that the table you are given goes. For x you get -2 -1 0 1 2 and for y you would get 16 4 1 1/4 1/16.

You could try y = (1/4)^x
For this try, you would get x = -2 -1 0 1 2 and for y = 16 4 0 1/4 and 1/16 
but that doesn't work either.

You could try until you get y = 4^x which does work.

g(x) is a lot easier to deal with. It looks better behaved. as x goes up, so does y. You will find that the y values obey y = x + 4. You could try other lines, but that one works. Many times it's just a guess

7 0
3 years ago
Please help i need to answer this within 2 hours
Gelneren [198K]
The answer is A. -137.8
7 0
3 years ago
Read 2 more answers
1. Petra tenía 49 canicas azules y dos canicas rojas. ¿Cuántas canicas deberá quitar Petra para qu
Mazyrski [523]

Answer:

Me no espanol

Step-by-step explanation:

LOL

6 0
3 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
A Shopkeeper
Sergio [31]

Answer:

The cost of bicycle is Rs1000

Step-by-step explanation:

Let x be the cost of bicycle

Total profit gained by the shopkeeper by selling at labelled price = 20%

Suppose he sell the bicycles at at 5% discount

Which means that the total profit he learn after discount will be:

20% -5% = 15%

The shopkeeper earns a profit of 15% if he sells at discount.

Profit gained by shopkeeper is:

15% of x = 15/100 · x

15% of x = 0.15x

Thus the profit gained will be 0.15x. As profit gained is equal to 150, we can say that

0.15x = 150

x = 1000

The cost of bicycle is Rs1000

4 0
3 years ago
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