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Semenov [28]
3 years ago
15

What's the solution to this inequality?

Mathematics
2 answers:
dezoksy [38]3 years ago
8 0
P< -9 IS THE ANSWER
  _
I'm trying to write, less than and equal to
garik1379 [7]3 years ago
6 0
P is less than or equal to -9
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If f(x) = -3x - 5 and g(x) = 4x - 2, find (f+ g)(x).
STALIN [3.7K]

Answer:

-40x

Step-by-step explanation:

f(x) =-3x-5

g(x) =4x-2

(f+g)(x) =?

now,

f(g(x))

=f(4x-2)

=-3×4x-2-5

=-10×4x

=-40x

7 0
2 years ago
Find the measure of x and y (NO LINKS OR REPORT)
Alchen [17]

Answer:

y= 35

x=55

Step-by-step explanation:

   

8 0
3 years ago
1.) Maggie has a collection of 1,200 Pennies. Of these, 25% are dated before 1980, 35% are dated from 1980-
Strike441 [17]

Answer:

rest of them (40%)

Step-by-step explanation:

4 0
3 years ago
Given the function ƒ(x )=x^2 - 5, find ƒ(7).<br><br> 49 <br><br> 2 <br><br> 44
rosijanka [135]
Substitute the 7 in anywhere there is an x.
49-5=44
4 0
3 years ago
Read 2 more answers
Find a function where f(0)=2 and f(1)=2
xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

f(x) = 2 is a valid solution.

Want something more fun? Why not a parabola? f(x)= ax^2+bx+c.

At this point you have three parameters to play with, and from the fact that f(0)=2 we can already fix one of them, in particular c=2. At this point I would recommend picking an easy value for one of the two, let's say a= 1 (or even a=-1, it will just flip everything upside down) and find out b accordingly:f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1

Our function becomes

f(x) = x^2-x+2

Notice that it works even by switching sign in the first two terms: f(x) = -x^2+x+2

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: f(x) = A cos (kx)

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need A= 2, and at that point the first condition is guaranteed; using the second to find k we get 2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi

f(x) = 2cos(2\pi x)

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

f(x)= 2+sin(2\pi x)

Sky is the limit.

8 0
3 years ago
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