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3 years ago

Who has insta grammm because I need help with my ela assignments and I need a good grade on them

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

7 0

Answer:

Happy new year to you and your family and

Feliz [49]3 years ago

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Me
May I please have brainliest?

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lilavasa [31]

Part A the second measurement is 1 because everything you do to the numerator you do to the denominator. So for Part A the other cups is 1

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4 years ago

3 times (square root of 125 over 5 minus 8)

fenix001 [56]

Hope this helps you good luck

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3 years ago

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A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

p is a polynomial of degree 5. p has roots of multiplicity 2 at t=4 and t=0, a root of multiplicity 1 at t=-4, and p(1)=2025. Fi

DENIUS [597]

P(x) = (x^2)(x - 4)^2(x + 4) + some constant(b)
2025 = (1^2)(1 - 4)^2(1 + 4) + b
2025 = 45 + b
b = 1980

Complete Equation:

p(x) = (x^2)(x - 4)^2(x +4) + 1980

or expanded form

p(x) = x^5 - 4x^4 - 16x^3 + 64x^2 + 1980

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3 years ago

Find each sum use the symbols to tell if the sum is greater than or less than

malfutka [58]

no offence but how r we supposed\ to help me when there is nothing to look at bro

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3 years ago

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