All categories

3 years ago

Please help meeee I’ll do anything

Mathematics

2 answers:

Anna11 [10]3 years ago

5 0

Answer:

B or second option

Step-by-step explanation:

The cost is the y-axis and the minutes is the x-axis. The + 2.00 is where the intersection of the y-axis starts.

FinnZ [79.3K]3 years ago

5 0

Answer: b is most likely to be it

Step-by-step explanation:

You might be interested in

Find the surface area of the rectangular prism. 4 mi 5 mi 4 mi

puteri [66]

Answer:

112

Step-by-step explanation:

A = 2 (wl+hl+hw)

A = 2 (4 x 5 + 4 x 5 + 4 x 4)

A = 2 (20 + 20 + 16)

A = 2 x 56

A = 112

5 0

3 years ago

Without using a calculator how do you evaluate tan 5π/6.

Mama L [17]

Remember that pi=180,
so 180/6=30, 30*5=150,
what's tan 150?
Remember that 150 degrees is in the 2nd quadrant, and this is tangent, so it means the answer must be negative, because tan 150= - tan 30, and -tan30=-√3/3.

Answer: -√3/3.

3 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

What is the median of the data? 3 8 5 5.5

ladessa [460]

Answer:

Median= middle number

(Example) 1, 4 ,6, 7 ,12

6 Would be the median here

so, 3, 3, 4, 4, 5, 5, 5, 5, 6, 7, 8,

5 Is the middle number and the median

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Help pls lol its timed

zaharov [31]

The total area of the shape is 32

6 0

3 years ago

Read 2 more answers