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Wittaler [7]
3 years ago
9

HURRY 25 POINTS IF RIGHT

Mathematics
2 answers:
VashaNatasha [74]3 years ago
6 0

Answer:  160

Step-by-step explanation:

The fuctoin that demenstrates this problem is 32=24+0.05x so all you have to do is solve for xyou would subtract 24 from bolth sides geting 8=0.05x then divide bolth sidews by o.o5 wich you get 160=x so the answer is 160.

Elina [12.6K]3 years ago
4 0

Answer:

A) 160 minutes

Step-by-step explanation:

First we write an equation

0.05x + 24 = 32

Subtract 24 from both sides

0.05x=8

Divide by 0.05 on both sides

x=160

So that mean at 160 minutes the plans will cost the same!

<em>Please mark brainliest! :)</em>

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Given a function f(x)=2x^2+3, what is the average rate of change of f on the interval [2, 2+h]?
-Dominant- [34]
-8h - 2h^2

f(2)-f(2+h) = -8h - 2h^2


Mark brainliest please
3 0
2 years ago
The cone and the cylinder below have equal surface area. True or false.
WARRIOR [948]

Answer:

False

Step-by-step explanation:

The surface area of the cone is

V=\pi r^2 +\pi rl

SA=\pi r^2 +\pi\times r\times 2r

SA=\pi r^2 +2\pi\times r^2

SA=3\pi r^2

The surface area of the cylinder is:

SA=2\pi r^2 +2\pi rh

SA=2\pi r^2 +2\pi\times r\times r

SA=4\pi r^2

3 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
The diagram shows a square.
frosja888 [35]

Answer:

3.5 cm

Step-by-step explanation:

6x -1 = 4x + 6

2x = 7

x = 3.5

6 0
3 years ago
3,4,4,5,5,6,6,6,6,7,9,12,14,15 Min: Q1: Med: Q3: Max:
kolezko [41]

Answer:

Median ( Q2) = 6

Q1 ( lower quartile) = 4.5

Q3 ( upper quaartile) = 10.5

Step-by-step explanation:

<u>3,4,4,5,5,6,  </u>6,6  ,<u>6,7,9,12,14,15</u>

6,6 is the median but we only need I median. So 6 + 6 = 12, 12 divided 2 = 6.

3,4,<u>4,5,</u>5,6,  is the Q1 ( lower quartile ) and median of Q1, is 4,5. 4 + 5 = 9. 9 divided 2 = 4.5

6,7,<u>9,12,</u>14,15  is Q3 ( upper quartile) 9,12 is median but we only need one. So 9 + 12 =21. 21 divided 2 = 10.5

5 0
2 years ago
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