Carla use two and three cups and one and three eight how much does she use in all

Please help and explain to me this. Ive been having difficulty in this

Need points sorry but hope u get the right answer

8 0

2 years ago

A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. The followin

dezoksy [38]

Answer:

32 days

Explanation:

Simulate (build a table) the growing of the number of pennies for some nights to figure out the pattern:

First night: 1 penny = 2⁰

Second night: 1 × 2 pennies = 2¹

Third night: 2 × 2 = 2²

Fourth nigth: 2² × 2 = 2³

nth night: 2ⁿ⁻¹

You want 2ⁿ⁻¹ ≥ 2,000,000,000

Which you solve in this way:

2ⁿ⁻¹ ≥ 2,000,000,000

2ⁿ⁻¹ ≥ 2,000,000,000

n-1 log (2) ≥ log (2,000,000,000)

n - 1 ≥ log (2,000,000,000) / log (2)

n - 1 ≥ 30.9

n ≥ 31.9

Since n is number of days, it is an integer number, so n ≥ 32.

Hence, she will have a total of more than $ 2 billion after 32 days.

You can prove that by calculating 2³² = 2,147,483,648.

5 0

3 years ago

Which of ordered pairs is not a function ?

Free_Kalibri [48]

$Answer:4)\\\\because\\(2;\ 3)\to x=2\ and\ y=3\\(2;\ 4)\to x=2\ and\ y=4\\\\for\ this\ same\ "x"\ y=3\ and\ y=4\\\\and\\\\(4;\ 5)\to x=4\ and\ y=5\\(4;\ 6)\to x=4\ and\ y=6\\\\for\ this\ same\ "x"\ y=5\ and\ y=6$

4 0

3 years ago

Show me how you solve it

julia-pushkina [17]

Answer:

5^20

Step-by-step explanation:

Law of Exponent I

$\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} }$

Therefore:

$\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{8 - 3})^{4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{5})^{4} }$

Law of Exponent II

$\displaystyle \large{( {a}^{m} ) ^{n} = {a}^{m \times n} }$

Thus:

$\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{8 - 3})^{4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{5})^{4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = {5}^{5 \times 4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = {5}^{20} }$

7 0

2 years ago

A tank contains 100 L of water. A solution with a salt con- centration of 0.4 kg/L is added at a rate of 5 L/min. The solution i

Fantom [35]

Answer:

a) (dy/dt) = 2 - [3y/(100 + 2t)]

b) The solved differential equation gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time

The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)

The rate of change of the volume of solution = dV/dt

Rate of flow into the tank = Fᵢ = 5 L/min

Rate of flow out of the tank = F = 3 L/min

(dV/dt) = Fᵢ - F

(dV/dt) = (Fᵢ - F)

dV = (Fᵢ - F) dt

∫ dV = ∫ (Fᵢ - F) dt

Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t

V - 100 = (Fᵢ - F)t

V = 100 + (5 - 3)t

V = 100 + (2) t

V = (100 + 2t) L

Component balance for the amount of salt in the tank.

Let the initial amount of salt in the tank be y₀ = 0 kg

Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min

Amount of salt in the tank, at any time = y kg

Concentration of salt in the tank at any time = (y/V) kg/L

Recall that V is the volume of water in the tank. V = 100 + 2t

Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min

Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)

(dy/dt) = 2 - (3y/V)

(dy/dt) = 2 - [3y/(100 + 2t)]

To solve this differential equation, it is done in the attached image to this question.

The solution of the differential equation is

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration after 20 minutes.

After 20 minutes, volume of water in tank will be

V(t) = 100 + 2t

V(20) = 100 + 2(20) = 140 L

Amount of salt in the tank after 20 minutes gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵

y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵

y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵

y(20) = 56 - 24.15 = 31.85 kg

Amount of salt in the tank after 20 minutes = 31.85 kg

Volume of water in the tank after 20 minutes = 140 L

Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L

Hope this Helps!!!

8 0

2 years ago