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sasho [114]
3 years ago
7

Factorization prima de 1024

Mathematics
1 answer:
Tamiku [17]3 years ago
3 0
1024
/ \
32 32
/ \ / \
8 4 8 4
/\ /\ /\ /\
(2) 4 (2) (2)(2) 4 (2) (2)
/\ /\
(2)(2) (2)(2)

2x2x2x2x2x2x2x2x2x2 or 2^10

*(There wasn’t a little 10 to use so I just put 2^10 which is two to the tenth power)

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How does replacing f(x) with f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative) affect
ivann1987 [24]
F(x) + k - Moves the graph k units up.

k f(x)   stretches the graph parallel to y-axis by a facor k

f (kx) stretches  the  graph by a factor 1/k parallel to x-axis

f(x + k)  moves the graph   3 units to the left.

For k negative the first one moves it k units down

for second transform negative does same transfoormation but also reflects the graph in the x axis

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4 0
3 years ago
What is the classification for this polynomial?
Mariana [72]

Answer:

Trinomial

Step-by-step explanation:

This is because there are three terms involved in the equation.

One of them is -7ab^4. Another is 4c^3. The third one is 25.

4 0
3 years ago
What two numbers add to +5 and have a product of -3?
Tasya [4]
They don't come out even.
As rounded decimals, the two numbers are

<em>5.54138...</em>  and  <em>-0.54138...</em>


6 0
3 years ago
Is each number correctly expressed in scientific notation? Choose yes or no​
Andre45 [30]
Answer : yes



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8 0
3 years ago
An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same
elena-s [515]

Answer:

Approximately 0.15 (360 / 2401.) (Assume that the choices of the 5 passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all 7 floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these 5 passenger could choose from any one of the 7 floors. There would be a total of 7 \times 7 \times 7 \times 7 \times 7 = 7^{5} unique ways for these 5\! passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the 7 floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only (7 - 1) = 6 floors.

Likewise, the third passenger would have to choose from only (7 - 2) = 5 floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only (7 \times 6 \times 5 \times 4 \times 3) unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the 7 floors. Each of these 7^{5} combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}.

5 0
2 years ago
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