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Andrew [12]
2 years ago
10

In graph, the area below f(x) is shaded and labeled A, the area below g(x) is shaded ad labeled B, and the area where f(x) and g

(x) have shading in common is labeled AB
the graph represents which system of inequalities?

y≤-2x+3
y≤x+3

y≥-2x+3
y≥x+3

y≤-3x+2
y≤-x+2

y>-2x+3
y>x+3​​​

Mathematics
1 answer:
NNADVOKAT [17]2 years ago
5 0

Answer:

The graph represents the system of inequalities y ≤ -3x + 2 and y ≤ -x + 2 ⇒ C

Step-by-step explanation:

From the given figure

∵ The direction of each line is to left

∴ The slopes of the lines are negative

∵ The slope of the line is the coefficient of x

∴ The coefficient of x in each inequality is negative ⇒ (1)

∵ The two lines intersect the y-axis at the point (0, 2)

∴ The y-intercept of the two lines is (0, 2)

∵ The y-intercept is the numerical term in the equation

∴ The numerical term in each inequality is 2 ⇒ (2)

∵ The two lines are solids

∵ The shaded area of each one is under the line

∴ The sign of inequality in both equations is ≤ ⇒ (3)

→ Look at the answer and find which one has the 3 conditions above

∵ y ≤ -3x + 2 and y ≤ -x + 2

∴ m = -3 and m = -1 ⇒ negative coefficients of x

∴ b = 2 ⇒ numerical term

∵ The sign of inequality is ≤

∴ The graph represents the system of inequalities y ≤ -3x + 2 and y ≤ -x + 2

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Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

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We are given the following function:

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v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42

Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

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