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BARSIC [14]
1 year ago
13

Give an example of a polynomial function that meets the above criteria

Mathematics
1 answer:
Komok [63]1 year ago
7 0

Answer:

F(x) = -5x - 8

Explanation:

Let's try with the polynomial function

f(x) = -5x - 8

We can calculate f(8) replacing x by 8, so:

f(8) = -5(8) - 8

f(8) = -40 - 8

f(8) = -48

Additionally, f'(x) will be equal to:

f'(x) = -5

And -5 is distinct from 0.

Finally, if f'(x) = -5, f''(x) is the derivative of a constant, so:

f''(x) = 0 for all x

Therefore, an function that satisfies all the conditions is:

F(x) = -5x - 8

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Please help me, this is my last question and I’ve been trying all day
aalyn [17]

Answer:

sqrt(x-3)+3

Step-by-step explanation:

Consider the graph of a function f(x), if we want to move the graph right by a units we have f(x-a), to then move it up by b units we have f(x-a)+b. In this case you replace the x with x-a to get sqrt(x-a)+b

Notice how the graph goes through (3,3) and (4,4) like the original graph goes through (0,0) and (1,1). This means we want to translate it 3 up and 3 right giving us the answer.

4 0
2 years ago
5, minus, start fraction, t, divided by, 3, end fraction when t=12
11Alexandr11 [23.1K]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow Final  \:\:Value = 1\:

____________________________________

\large \tt Solution  \: :

The statement can be represented as :

\qquad \tt \rightarrow \: 5 -  \dfrac{t}{3}

[ for t = 12 ]

\qquad \tt \rightarrow \: 5 -  \dfrac{12}{3}

\qquad \tt \rightarrow \: 5 - 4

\qquad \tt \rightarrow \: 1

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

7 0
2 years ago
Read 2 more answers
How many two-digit numbers are there in which the tens digit is greater than<br> the ones digit?
Dovator [93]

there are 46 2-digit numbers in which the tens digit is greater than the ones digit.

<h3>How many 2-digit numbers are there in which the tens digit is larger than the ones digit?</h3>

A 2-digit number is written as:

a*10 + b

Where a and b are single-digit numbers.

Tens digit and ones digit refers to place values:

a is the tens digit, and b is the ones digit. The numbers with a larger tens digit than the ones digit are:

98, 97, 96, 95, 94, 93, 92, 91, 90, 87, 86, 85, 84, 83, 82, 81, 80, 76, 75, 75, 74, 73, 72, 71, 70, 65, 64, 63, 62, 61, 60, 54, 53, 52, 51, 50, 43, 42, 41, 40, 32, 31, 30, 21, 20, 10.

We can see that there are 46 numbers in that list, so we conclude that there are 46 2-digit numbers in which the tens digit is greater than the ones digit.

If you want to learn more about place values:

brainly.com/question/12386995

#SPJ1

7 0
1 year ago
Please help me prove this theorem thing. I'll give yuh mind kisses &lt;3 have a good day luvlies
lesantik [10]

Answer:

∠ 1 = 20 x + 5

∠ 2 = 24 x - 1

If lines l and m are parallel:

∠ 1 + ∠ 2 = 180°

20 x + 5 + 24 x - 1 = 180

44 x + 4 = 180

44 x = 180 - 4

44 x = 176

x = 176 : 44

x = 4°

Your welcome again :)

Step-by-step explanation:

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=g%28x%29%20%3D%203x%20-%206" id="TexFormula1" title="g(x) = 3x - 6" alt="g(x) = 3x - 6" align=
VARVARA [1.3K]

Hey there! :)

Answer:

g^{-1} (g(10))=10

Step-by-step explanation:

Begin by calculating g(10):

g(x) = 3x - 6

Substitute in 10 for x:

g(10) = 3(10) - 6

g(10) = 30-6

g(10) = 24.

Plug '24' into 'x' into g^{-1} (x)

g^{-1} (24)=\frac{(24) + 6}{3}

Simplify:

g^{-1} (24)=\frac{30}{3}

g^{-1} (24)=10

3 0
3 years ago
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