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Sidana [21]
3 years ago
7

What is the value of the function f(x) = 2x – 5 when x = 22? 17 39 49 105

Mathematics
2 answers:
prohojiy [21]3 years ago
5 0
Take the original equation, substitute x for 22, and use PEMDAS to solve.

f(x) = 2x - 5
f(x) = 2(22) - 5
f(x) = 44 - 5
f(x) = 39

Your final answer will be 39.
vlabodo [156]3 years ago
4 0
Substitute 22 for x
2(22)-5
44-5
39
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Find the GCF from the two numbers and rewrite the sum using nthe distributive property
Vadim26 [7]

Answer:

The greatest common factor is 6.

Step-by-step explanation:

Greatest common factor is 6. If you use the distributive property then the answer would be 6(4) + 6(6) or 6(4+6). Then you distribute the 6 to each digit and should get 24+36.

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2 years ago
What is the midpoint ?
inn [45]

Answer:

Step-by-step explanation:

A(-5,-4)  B(-3, 3)

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(-4 + 3)/2 = -1/2

(-4, -1/2) the midpoint

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3 years ago
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

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If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
Subtract -2k^3 + k^2 - 9 from 5k^3 - 3k + 7.
Lady_Fox [76]
5k^3 - 3k + 7 + <span>2k^3 - k^2 + 9 
= 7k^3 - k^2 -3k + 16</span>
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3 years ago
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Katyanochek1 [597]
I don't have an explanation but I have an answer 0.02
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3 years ago
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