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alina1380 [7]
3 years ago
11

Consider the equation 1/3x−2=x+2. Write a system of linear equations using each side of the equation. Then solve the original eq

uation by graphing the system of linear equations.
Then graph and solve for x
Mathematics
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:

(-6,4)

Step-by-step explanation:

1/3x − 2 = x + 2 is obtained by equating the following two functions:

y = (⅓)x - 2

y = x + 2

Graph:

Draw two straight lines

1) joining (0,-2) and (6,0)

2) joining (0,2) and (-2,0)

On the graph, you'll see that the above two lines intersect at: (-6,4)

Hence the solution is (-6,4)

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Factoring / gcf ??? (in picture)
Licemer1 [7]

Step-by-step explanation:

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6 0
2 years ago
Noor spends Rs.1440 out of Rs.2000 and saves the remaining find the percentage of his savings​
matrenka [14]

Answer:

28 percent............

3 0
3 years ago
What is the expression here.. I found the answer but i dont know the expression!! SOMEONE PLEASE HELP!!
iogann1982 [59]
Idk if this helped you

5 0
3 years ago
Add:<br> 21.4 + (- 23.9)
pogonyaev
The answer is -2.5. this is how you work it out.
21.4 + (-23.9) = -5/2 = -2.5

4 0
3 years ago
How do I do question 9? Please give answer thank you
kvasek [131]
<h3>Given</h3>
  • a cone of height 0.4 m and diameter 0.3 m
  • filling at the rate 0.004 m³/s
  • fill height of 0.2 m at the time of interest
<h3>Find</h3>
  • the rate of change of fill height at the time of interest
<h3>Solution</h3>

The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of

... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²

This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.

... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s

Dividing by the coefficient of dh/dt, we get

... dh/dt = 0.004·16/(0.09π) m/s

... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s

_____

You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)

Note that the cone dimensions mean the radius is 3/8 of the height.

V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³

dV/dt = 9π/64·h²·dh/dt

.004 = 9π/64·0.2²·dh/dt . . . substitute the given values

dh/dt = .004·64/(.04·9·π) = 32/(45π)

7 0
3 years ago
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