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SpyIntel [72]
2 years ago
7

If f(x)=3x-1, find f(f(x) and f(f(f(x)))

Mathematics
1 answer:
Arisa [49]2 years ago
8 0

Answer:

i hope this helps

Step-by-step explanation:

you just need to change the x to fx

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What is the value of y if log <base 5> <index -7> = y?​
Dvinal [7]

Answer:

Hence, the value of y is undefined.

Step-by-step explanation

log_{5} (-7) = \frac{log(-7)}{log(5)} from the property, log_a (b) = (log(a))/log(b).

The value log (-7) was not defined. So, the entire log value also not defined.

Hence, the value of y is undefined.

6 0
2 years ago
NEED HELP ASAP ITS TIMED
Arturiano [62]
B. 39

Good luck on yourself timed exam!
4 0
3 years ago
TIMED
Rudik [331]
-x^2 + 4x + 12 = -3x + 24

-> x^2 - 7x + 12 = 0

-> (x-3)(x-4) = 0

-> x= 3 or 4

so y = 15 when x = 3, y = 12 when x = 4
6 0
3 years ago
Read 2 more answers
Solve the given equation for v.
vaieri [72.5K]

Answer:(3t)^2

Step-by-step explanation:

Steps below

4 0
3 years ago
I really really really need help!!!!
Yakvenalex [24]
For f to be continuous at x=1, you need to have the limit from either side as x\to1 to be the same.

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b

If a=2 and b=3, then the limit from the right would be 2+3=5\neq2, so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation a+b=2.

Part (c) is done similarly to part (1). This time, you need to limits from either side as x\to2 to match. You have

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b
\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0

So, a and b have to satisfy the relation 4a+2b=0, or 2a+b=0.

Part (4) is done by solving the system of equations above for a and b. I'll leave that to you, as well as part (5) since that's just drawing your findings.
8 0
3 years ago
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