Ask question

Ask question

All categories

3 years ago

11

Please help me, or I will fail.

1 answer:

Contact [7]3 years ago

6 0

Answer:

$\frac{1}{4} +m=\frac{2}{3}$

Step-by-step explanation:

The tape diagram is showing that by adding a length 'm' to a length of $\frac{1}{4}$ , then you will get a length of $\frac{2}{3}$ .

You might be interested in

A geologist collected a sample of quartz weighing 3.554 kilograms. How does the value represented by the 5 in the tenths place c

umka2103 [35]

Five tenths is forty-five hundredths greater than 5 hundredths

5 0

3 years ago

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

Find the value of 3<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D" id="TexFormula1" title="\sqrt{x}" alt="\sqrt{x}" align="abs

Alex

Answer:

The answer is

<h2>22</h2>

Step-by-step explanation:

3√x + 7

x = 25

Substitute the value of x into the expression

That's

3(√25) + 7

√25 = 5

So we have

3(5) + 7

= 15 + 7

<h3>= 22</h3>

Hope this helps you

8 0

4 years ago

What is the value of 2x^3+4x^2-6x, where x= 3? A. 63 B.45 C. 35 D. 99

Sunny_sXe [5.5K]

2(3)^3 + 4(3)^2-6(3)
2x 27 + 4x9 - 18
54 + 36 -18
= 72

5 0

4 years ago

I do not know the answer

Vaselesa [24]

Answer:

5 1 over 4

Step-by-step explanation:

6 0

4 years ago