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3 years ago

Warm-Up

Mathematics

1 answer:

seraphim [82]3 years ago

4 0

Answer:

$y=2\frac{1}{6}$ or $\frac{13}{6}$

Step-by-step explanation:

We know that $x$ and $y$ are directly proportional. Therefore, using the given information, we can set up the following proportion:

$\frac{6.5}{12}=\frac{y}{4}$

Solving for $y$ , we get:

$\frac{6.5}{12}=\frac{y}{4}$

$12*y=6.5*4$ (Cross-Products Property)

$12y=26$ (Simplify)

$\frac{12y}{12} =\frac{26}{12}$ (Divide both sides of the equation by $12$ to get rid of $y$ 's coefficient)

$y=2\frac{1}{6}$ or $\frac{13}{6}$ (Simplify)

Hope this helps!

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Given a normal distribution with a mean of 125 and a standard deviation of 14, what percentage of values is within the interval

Anit [1.1K]

Answer:68%

Step-by-step explanation:

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4 years ago

Read 2 more answers

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

Can someone please help me with math.

zlopas [31]

Answer:

the answer to the questions are c,b,a

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3 years ago

Can someone help me ?

Darina [25.2K]

Answer:

where ever your at looks dark.

Step-by-step explanation:

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3 years ago

A trapezoid has an area of 22 square inches. The longer base measures 6 inches and the height is 4 inches. use the formula <img

AleksandrR [38]

22=1/2*4(6+b)
22=2(6+b)
22=12+2b
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Divide both sides by 2:
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The answer is b.

Hope this helps :)

3 0

4 years ago