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adelina 88 [10]
3 years ago
15

Maxine was given the following expression: 3(40+ n).

Mathematics
2 answers:
Cloud [144]3 years ago
8 0
Answer: the expressions above has 2 factors. 3 and 40 + n
GREYUIT [131]3 years ago
7 0

Answer: RICK ROLLED YOU HAHA

Step-by-step explanation: 3 and 40 + n

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Sin x =cos 19 X =sin^1(cos 19) X=71
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Billy is playing a game using a bag with 11 marbles inside. The bag contains 3 red, 3 orange, 1 yellow, 2 purple marbles, and 2
Dmitrij [34]

Answer: 3/55

Step-by-step explanation:

From the information given, the bag contains 3 red, 3 orange, 1 yellow, 2 purple marbles, and 2 Pink marbles. Each time he picks an orange marble, she will win a prize.

If he picks a marble the first time, the probability of picking an orange marble will be 3/11. After that we will have 10 marbles left as one has been picked and have 2 orange marbles left, then the probability of picking another orange marble will be 2/10.

Therefore, the probability he will win a prize on both picks will be:

= 3/11 × 2/10

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= 3/55

4 0
3 years ago
A line segment has endpoints P(-1,5)and Q(5,2) find the coordinate of that trisect the segment​
musickatia [10]

Answer:

(1, 6).  The second is (-1 + 2 + 2, 1 + 1 + 1), or (3, 3)

Step-by-step explanation:

Moving from P to Q, we see that x increases by 6  and y increases by 3.

"Trisect" means "divide into three equal subintervals."

We find (1/3) of 6 (which comes out to 2) and (1/3) of 3 (which comes out to 1).  

Thus, the first junction (between 1st and 2nd trisection) is (-1 + 2, 5 + 1), or (1, 6).  The second is (-1 + 2 + 2, 1 + 1 + 1), or (3, 3)

5 0
3 years ago
From first principles, find the indicated derivatives​
LenaWriter [7]

By definition of the derivative,

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)

\displaystyle\frac{dr}{ds} = \frac{3s^2}2

6 0
2 years ago
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