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natulia [17]
3 years ago
10

A liner function of x is graphed on a coordinate grid. the points 6,34 and 18,26 lie on the graph of the function what are the r

ate of change of the function and the value of the function the x=24
please show or explain the work so l can understand
Mathematics
1 answer:
Lynna [10]3 years ago
4 0

Answer:

Rate of change: m = -2/3

Value of the function at x = 24: y = 22

Step-by-step explanation:

You can write linear functions in <u>slope-intercept form</u>: y = mx + b

"x" and "y" are points on the line

"m" is the slope, how steep the line is, also known as <u>rate of change</u>

"b" is the y-intercept, when the graph hits the y-axis

Find the rate of change of the function by finding the slope between the two given points.

The formula for slope: m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Label point 1 and point 2. Label each coordinate; remembered ordered pairs are (x, y).

Point 1      (6, 34)         x₁= 6   y₁ = 34

Point 2     (18, 26)        x₂ = 18   y₂ = 26

Substitute the points into your equation using the bolded information.

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

=\frac{26-34}{18-6}       Subtract

=\frac{-8}{12}             Reduce the fraction. Top and bottom divide by 4

=\frac{-2}3}              Slope of the line, "m"

Substitute the slope and any point you want into slope-intercept form. I choose point (6, 34). x = 6    y = 34   m = -2/3

y = mx + b

34 = (-2/3)(6) + b                      Multiply to simplify

34 = (-12/3) + b

Start isolating "b"

34 + 12/3 = (-12/3) + 12/3 + b              Add 12/3 on both sides

34 + 12/3 = b

Add fractions by combining the whole number into the numerator. Multiply the whole number by the denominator and add to the numerator.

b = (34*3 + 12) / 3              Solve numerator

b = 114/3                      Value of "b"

Substitute the values "m" and "b" into slope-intercept form.

y = -2/3 x + 114/3                  Equation of the line

To find the x = 24, take the equation of the line and substitute "x" for 24. Then solve for "y".

y = -2/3 x + 114/3                       Equation of the line

y = (-2/3)(24) + 114/3

y = (-2*24)/3  + 114/3                     Multiply into the numerator

y = (-48)/3 + 114/3                       Combine under the same denominator

y = (-48 + 114) / 3                      Add

y = 66/3                        Divide

y = 22                      Value of function

Therefore the rate of change, or slope "m", is -2/3. When "x" is 24, the value of the entire function is 22.

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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.

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3 years ago
pls pls pls help i give brainest and i will give 37 points and i will give a thanks plssss sove 22 and 23
Alex787 [66]
<h2>Answer:</h2><h2>what do u mean solve 22&23</h2>

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Answer:

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Step-by-step explanation:

Given:

g(x)=4x^2+2x\\ \\f(x)=\int\limits^x_0 {g(t)} \, dt

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f(6)

First, find f(x):

f(x)\\ \\=\int\limits^x_0 {g(t)} \, dt\\ \\=\int\limits^x_0 {(4t^2+2t)} \, dt\\ \\=\left(4\cdot \dfrac{t^3}{3}+2\cdot \dfrac{t^2}{2}\right)\big|\limits^x_0\\ \\=\left(\dfrac{4t^3}{3}+t^2\right)\big|\limits^x_0\\ \\= \left(\dfrac{4x^3}{3}+x^2\right)-\left(\dfrac{4\cdot 0^3}{3}+0^2\right)\\ \\=\dfrac{4x^3}{3}+x^2

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