Any suggestions for a word problem for the equation : 5/12 * 1/6?

How many milliliters is in 50 liters of gas

mojhsa [17]

50000 millilitres Is in 50 liters of gas.

8 0

3 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 4 (a) Find the interval on which

kozerog [31]

Answer:

a) The function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Written in interval form

(-∞, -1.45) and (3.45, ∞)

- The function, f(x) is decreasing at the interval (-1.45 < x < 3.45)

(-1.45, 3.45)

b) Local minimum value of f(x) = -78.1, occurring at x = 3.45

Local maximum value of f(x) = 10.1, occurring at x = -1.45

c) Inflection point = (x, y) = (1, -16)

Interval where the function is concave up

= (x > 1), written in interval form, (1, ∞)

Interval where the function is concave down

= (x < 1), written in interval form, (-∞, 1)

Step-by-step explanation:

f(x) = x³ - 6x² - 15x + 4

a) Find the interval on which f is increasing.

A function is said to be increasing in any interval where f'(x) > 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

the function is increasing at the points where

f'(x) = 3x² - 6x - 15 > 0

x² - 2x - 5 > 0

(x - 3.45)(x + 1.45) > 0

we then do the inequality check to see which intervals where f'(x) is greater than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

So, the function (x - 3.45)(x + 1.45) is positive (+ve) at the intervals (x < -1.45) and (x > 3.45).

Hence, the function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Find the interval on which f is decreasing.

At the interval where f(x) is decreasing, f'(x) < 0

from above,

f'(x) = 3x² - 6x - 15

the function is decreasing at the points where

f'(x) = 3x² - 6x - 15 < 0

x² - 2x - 5 < 0

(x - 3.45)(x + 1.45) < 0

With the similar inequality check for where f'(x) is less than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

Hence, the function, f(x) is decreasing at the intervals (-1.45 < x < 3.45)

b) Find the local minimum and maximum values of f.

For the local maximum and minimum points,

f'(x) = 0

but f"(x) < 0 for a local maximum

And f"(x) > 0 for a local minimum

From (a) above

f'(x) = 3x² - 6x - 15

f'(x) = 3x² - 6x - 15 = 0

(x - 3.45)(x + 1.45) = 0

x = 3.45 or x = -1.45

To now investigate the points that corresponds to a minimum and a maximum point, we need f"(x)

f"(x) = 6x - 6

At x = -1.45,

f"(x) = (6×-1.45) - 6 = -14.7 < 0

Hence, x = -1.45 corresponds to a maximum point

At x = 3.45

f"(x) = (6×3.45) - 6 = 14.7 > 0

Hence, x = 3.45 corresponds to a minimum point.

So, at minimum point, x = 3.45

f(x) = x³ - 6x² - 15x + 4

f(3.45) = 3.45³ - 6(3.45²) - 15(3.45) + 4

= -78.101375 = -78.1

At maximum point, x = -1.45

f(x) = x³ - 6x² - 15x + 4

f(-1.45) = (-1.45)³ - 6(-1.45)² - 15(-1.45) + 4

= 10.086375 = 10.1

c) Find the inflection point.

The inflection point is the point where the curve changes from concave up to concave down and vice versa.

This occurs at the point f"(x) = 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

f"(x) = 6x - 6

At inflection point, f"(x) = 0

f"(x) = 6x - 6 = 0

6x = 6

x = 1

At this point where x = 1, f(x) will be

f(x) = x³ - 6x² - 15x + 4

f(1) = 1³ - 6(1²) - 15(1) + 4 = -16

Hence, the inflection point is at (x, y) = (1, -16)

- Find the interval on which f is concave up.

The curve is said to be concave up when on a given interval, the graph of the function always lies above its tangent lines on that interval. In other words, if you draw a tangent line at any given point, then the graph seems to curve upwards, away from the line.

At the interval where the curve is concave up, f"(x) > 0

f"(x) = 6x - 6 > 0

6x > 6

x > 1

- Find the interval on which f is concave down.

A curve/function is said to be concave down on an interval if, on that interval, the graph of the function always lies below its tangent lines on that interval. That is the graph seems to curve downwards, away from its tangent line at any given point.

At the interval where the curve is concave down, f"(x) < 0

f"(x) = 6x - 6 < 0

6x < 6

x < 1

Hope this Helps!!!

5 0

3 years ago

Evaluate 6+x6+x6, plus, x when x=3?

vampirchik [111]

Answer:

42

Step-by-step explanation:

x=3

6+x6+x6

6+(3)6+(3)6

6=6

(3)6=18

6+18+18

24+18

42

Hope this helps ;) ❤❤❤

8 0

3 years ago

Read 2 more answers

Can anyone help me solve this

yawa3891 [41]

Answer:

a. $45.54; b. $51.42; c. See below

Step-by-step explanation:

a. 193 min

Service charge = $40.00

1st 100 min @ 0.021 = 2.10

Next 93 min @0.037 = 3.44

Total = $45.54

b. 317 min

Service charge = $40.00

1st 100 min @ 0.021 = 2.10

2nd 100 min @ 0.037 = 3.70

117 min @ 0.048 = 5.62

Total = $51.42

c. Piecewise function

The charge is

$40.00 + 0.021t if t ≤ 100
$42.10 + 0.037(t - 100) if 100 < t ≤ 200
$45.80 + 0.048 (t - 200) if t > 200

which we can write like this:

$f(t) =\begin{cases}40.00 + 0.021t & t \leq 100\\42.10 + 0.037(t-100) & 100 < t \leq200\\45.80 + 0.048(t-200) & t > 200\end{cases}$

5 0

3 years ago

Let’s assume the following statements are true: Historically, 75% of the five-star football recruits in the nation go to univers

marishachu [46]

Given:
75% of the five-star football recruits in the nation go to universities in the three most competitive athletic conferences. → 25% goes to other schools.

five-star recruits get full football scholarships 93% of the time, regardless of which conference they go to. → 7% of the 5-star recruits don't get full football scholarships.

a. The probability that a randomly selected five-star recruit who chooses one of the best three conferences will be offered a full football scholarship?
75% * 93% = 69.75%

b. What are the odds a randomly selected five-star recruit will not select a university from one of the three best conferences?
25% of selected five-star recruit will not select a university from one of the three best conferences. I got the number based on the given data. Since, 75% will go, the remaining percent won't go. Total percentage should be 100% of the population.

c. Explain whether these are independent or dependent events. Are they Inclusive or exclusive?
These are independent events. One can still go to different school and still be legible for the full football scholarship.

For question 2, pls. see attachment.

4 0

3 years ago