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3 years ago

Which expression shows the first step in simplifying 2x – 3(x + 2y) – 5(y – 7x)?

Mathematics

2 answers:

artcher [175]3 years ago

5 0

Answer: Option 3.

Step-by-step explanation:

1. To simplify the expression shown in the problem, the first step is to apply the Distributive property, this means that you must multiply the number that are outside of the parentheses by the numbers inside of them.

2. Then, you have that the first step is:

$2x-3(x+2y)-5(y-7x)=2x-3x-6y-5y+35x$

3. Therefore, you can conclude that the correct answer is the third option.

viva [34]3 years ago

3 0

Answer:

2x - 3x - 6y -5y + 35x

Step-by-step explanation:

2x-3(x+2y)-5(y-7x)

We use order of operations PEMDAS

Lets start with parenthesis

We distribute -3 inside the parenthesis and Distribute -5 inside the parenthesis

2x - 3x - 6y -5y + 35x

combine like terms

-1x - 6y -5y + 35x

-1x - 11y + 35x

34x - 11y

first step is 2x - 3x - 6y -5y + 35x

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How do i solve an absolute prblem

pishuonlain [190]

Step 1: Isolate the absolute value expression.Step2: Set the quantity inside the absolute value notation equal to + and - the quantity on the other side of the equation.Step 3: Solve for the unknown in both equations.Step 4: Check your answer analytically or graphically.

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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.

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3 years ago

Mr. Scott’s workroom is a rectangle that measures 25 feet by 27 feet. His woodworking equipment takes up an area of 300 ft squar

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375 because if you do 27x25=675 then I did 675-300=375.
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