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grin007 [14]
4 years ago
8

Simplify the expression n^3 • 3n

Mathematics
1 answer:
N76 [4]4 years ago
4 0

Answer:

Step-by-step explanation:

3n x n^3 = 3n^4

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3/4 of Anna's books are equal to 1/3 of Jake's books. Find the ratio of the number of Anna's books to the number of Jake's books
irga5000 [103]
3/4=1/3 

cross multiply
7 0
3 years ago
Please help with 8.???
IRINA_888 [86]

Answer:

They are inverse functions

Step-by-step explanation:

A property of inverse functions is that if f(x) = a,  g(a) = x

We can plug in x = 3

f(1) = 3(3) - 2 = 7

That means that, supposing they are inverse functions, g(7) should equal 3

g(7) = \frac{7+2}{3} =3

It checks out

Another way to see if two functions are inverse is to swap the x and y of one of the functions.

ex. y = 3x-2\\-> x = 3y - 2\\-> x+2 = 3y\\-> y = \frac{x+2}{3}\\ \\

Since, after the swap, the functions are equal, we know it is an inverse function

5 0
3 years ago
Khan academy Help please
Aliun [14]

Answer:

B, C and F

Step-by-step explanation:

<u>A constant is where a variable or number is on its own</u>

B - 2x-4 (4 is the constant)

C - 8x+y (y is the constant)

F - 7+4x (7 is the constant)

7 0
3 years ago
Read 2 more answers
Which of the following expressions represents a function?
PSYCHO15rus [73]
A function can have one x-value for each y-value Or even, two difference x-values for one y-value.

And out of the four options, only A and C represent a function. The other two options have the same x value for two y-values.

One way to test if it is a function is by the vertical line test.
7 0
3 years ago
For each function f(n) and time t in the table, determine the largest size n of a problem that can be solved in time t, assuming
egoroff_w [7]
First recall that a microsecond is 10^−6 seconds. Hence, one second = 10^6 microseconds, one hour = 3600000000 = 3.6 • 10^9 microseconds, one month (assume a month has 30 days) = 2592000000000 = 2.592 • 10^12 microseconds, and one century = 3110400000000000 = 3.1104 • 10^15 microseconds.

Row 1: f(n) = log n In this case, we need to determine the largest n such that log n ≤ 1000000. To solve this inequality, we need to rewrite the inequality as 2^logn ≤ 2^1000000 or n ≤ 2^1000000. Recall from lecture that 2^10 ≈ 10^3, thus we have that 2^1000000 = 2^10•100000 = (2^10)^100000 ≈ (10^3)^100000 = 10^300000. This is the result given in the textbook. For one hour, we have that n ≤ 2^3600•1000000 and thus n ≈ 10^1080000000.

Row 8: f(n) = n! For us to see that the largest sized input is 12 that can be processed within an hour when f(n) = n! one can simply, compute 12! And verify that it is less than the number of microseconds in one hour, but that 13! is greater than the number of microseconds in an hour.

Row 4: f(n) = n log n In this case, use Maple to solve equations like n log n−1000000 = 0. The Maple command for solving this equation is fsolve(n*log[2](n) - 1000000 = 0)
8 0
4 years ago
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