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2 years ago

What is equivalent to 24b-34ab

Mathematics

1 answer:

jarptica [38.1K]2 years ago

6 0

This is what I got as the answer, 2b x (12-17a)

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Find a and b so that f(x) = x^3 + ax^2 + b will have a critical point at (2,3).

Digiron [165]

Using the critical point concept, it is found that a = -3 and b = 7.

<h3>What are the critical points of a function?</h3>

The critical points of a function are the values of x for which:

$f^{\prime}(x) = 0$

In this problem, the function is:

$f(x) = x^3 + ax^2 + b$

Hence, the derivative is:

$f^{\prime}(x) = 3x^2 + 2ax$

Then:

$f^{\prime}(x) = 0$

$3x^2 + 2ax = 0$

$x(3x + 2a) = 0$

$x = 0$

$3x + 2a = 0$

$3x = -2a$

$x = -\frac{2a}{3}$

Since the critical point is at x = 2, we have that:

$-\frac{2a}{3} = 2$

$-2a = 6$

$2a = -6$

$a = -3$

Then:

$f(x) = x^3 - 3x^2 + b$

Critical point at (2,3) means that when $x = 2, y = 3$ , then:

$3 = 2^3 - 3(2)^2 + b$

$8 - 12 + b = 3$

$b = 7$

You can learn more about the critical point concept at brainly.com/question/2256078

4 0

1 year ago

How many 1/3 inch cubes does it take to fill a box with a width of 3 1/3 inches, a length of 2 2/3 inches, and a height of 1 1/3

Charra [1.4K]

Answer:

It will take 3/4 cubes to fill the box

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Will mark brainliest

matrenka [14]

Answer: its 2.50 so B

Step-by-step explanation: I just did the test

6 0

2 years ago

Read 2 more answers

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F. Af

fgiga [73]

Answer:

53.3324

Step-by-step explanation:

given that a thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.

By Newton law of cooling we have

T(t) = $T+(T_0-T)e^{-kt}$

where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature

After half minute thermometer reads 60° F. Using this we can find k

$T(0,5) = 40+(70-40)e^{-k/2} = 60\\e^{-k/2} =2/3\\-k/2 = -0.4055\\k = 0.8110$

So equation is

$T(t) = 40+(30)e^{-0.8110t}\\$

When t=1,

we get

$T(1) = 40+(30)e^{-0.8110}\\\\=53.3324$

5 0

2 years ago

Help! This is due today and I'm stuck on the last problem.

FrozenT [24]

Answer:

Step-by-step explanation:

Square

519450.2

Holes

Semi-circle: 1/2*pi*r^2

diameter = 700/2 = 350 =radius

1/2*pi*350 = 175*pi

Triangle

1/2 b*h

1/2 200*400

100*400

40000

Take out the semi circle and triangle

560000-40000-175*pi

520000-175*pi = 519450.2

6 0

2 years ago