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ASHA 777 [7]
2 years ago
13

Aubrey and Charlie are driving to a city that is 120 miles from their house. They have already traveled 20 miles, and they are d

riving at a constant rate of 50 mi/h.
A function f models their distance from home as a function of time. Complete the statement below using the function you wrote in Item 4. ( the function in question 4 is f(x) = 50x + 20)

f(1.5) = ______
The value indicates that after_______
hours of driving, Aubrey and Charlie will be_______
miles from home.
Mathematics
1 answer:
mario62 [17]2 years ago
8 0

Answer:

f(1.5) = _95_

The value indicates that after_1.5 hours_

hours of driving, Aubrey and Charlie will be_95_

miles from home.

Step-by-step explanation:

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Which of the following is an equivalent form of the compound inequality −33 > −3x − 6 ≥ −6?
wolverine [178]

<u>ANSWER</u>

9\:   < x  \leqslant0

<u>EXPLANATION</u>

The given compound inequality is

- 33 \:  >  - 3x - 6 \geqslant  - 6

We need to simplify this inequality so that we can obtain x standing alone between the inequality signs.

We add 6 through out the inequality.

- 33  + 6\:  >  - 3x - 6 + 6 \geqslant  - 6 + 6

This simplifies to:

- 27\:  >  - 3x \geqslant  0

We now divide through by -3 and reverse the inequality sign.

\frac{- 27}{ - 3} \:   <   \frac{ - 3x}{ - 3}   \leqslant    \frac{0}{ - 3}

We now simplify to get:

9\:   < x  \leqslant    0

6 0
2 years ago
Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y
zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

So

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

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8 0
2 years ago
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