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Vesnalui [34]
2 years ago
11

I just got ace rank and I got 4k points, so to celebrate ill being giving away free brainiest and 100 points to Whoever guesses

the song right. Any bot answers will be deleted..
"30 to your face (YEAH) 60 to your face (YEAH)"
Mathematics
2 answers:
Shalnov [3]2 years ago
6 0

Answer:

XXL Freshmen

Step-by-step explanation:

Vitek1552 [10]2 years ago
5 0

Answer: Omg congratulations

Step-by-step explanation:

XXL Freshman i love that song

XXL Freshman by: Denzel Curry,Kodak Black,lil uzi vert and whatever his name is lol

Hold up, Rolex on my wrist

Rolex on my b** "what?", 30 on my waist

30 to ur face "yeah", 60 to ur face

90 to ur face "yeah", thats a closed case

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12 plus 7 divided by 2 help please
polet [3.4K]

Answer:

15.5

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Jill works on the weekends. Last saturday, she worked 5.75 hours. Sunday she worked 6.75 hours. she earn $7.50 per hours. How mu
leva [86]
93.75 because 5.75x7.50=43.125 and 6.75x7.50=50.625 and 50.625+43.125=93.748 and round it to the near test hundredth because it’s money and that’s 93.75
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2 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
What is the volume of the triangular Prism
baherus [9]
It is probably A. A is the most reasonable answer.
4 0
3 years ago
Help help help help help
Inessa [10]

Answer:

I believe it's 240°

Step-by-step explanation:

It's been awhile since I've done this but I believe that's the answer, here's a picture of a pug. Smile and have a good day!

5 0
2 years ago
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