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KonstantinChe [14]
3 years ago
12

Suppose that in the last few seconds you devoted to question 1 on your physics exam you earned 4 extra points, while in the last

few seconds you devoted to question 2 you earned 10 extra points. You earned a total of 48 and 12 points, respectively, on the two questions, and the total time you spent on each was the same. If you could take the exam again, how—if at all—should you reallocate your time between these questions?
Mathematics
1 answer:
slavikrds [6]3 years ago
4 0

Answer:

If you could take the exam again, you should spend more time solving the question 2.

Step-by-step explanation:

In the last few seconds of each question, you gained:

4 extra points in question 1. These points are a small fraction of your question 1 score.

10 extra points in question 2. These points are a large fraction of your question 2 score.

You gained 6 extra points in question 2 from the extra time. These 10 extra points are also a large fraction of the question 2 score, while your question 1 score was already good enough. This means that if you could take the exam again, you should spend more time solving the question 2.

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Now since he reads 35 pages per day.

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3 years ago
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Question 3
Sladkaya [172]

Answer:

(x+6)(x+2) is the factored equation

(-6, -2) is the coordinates

Step-by-step explanation:

find what multiplies to be 12 and adds to be 8

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3 years ago
The point (5/13,y) in the fourth quadrant corresponds to angle θ on the unit circle.
andreyandreev [35.5K]
Cos θ = adjacent / hypothenus = 5/13
opposite = sqrt(hypothenus^2 - adjacent^2) = sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12

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5 0
4 years ago
1. Evaluate the function f(x)=3x-1 using the domain of -2, 0, 3, and 5. Results are to be shown in a table.
IceJOKER [234]

Answer:

See explanation

Step-by-step explanation:

You are given the function f(x)=3x-1

The domain of the function are all possible values of variable x, so you can find f(x) for x=-2, 0, 3, 5:

f(-2)=3\cdot (-2)-1=-6-1=-7\\ \\f(0)=3\cdot 0-1=0-1=-1\\ \\f(3)=3\cdot 3-1=9-1=8\\ \\f(5)=3\cdot 5-1=15-1=14

The table is

\begin{array}{cc}x&f(x)\\-2&-7\\0&-1\\3&8\\5&14\end{array}

The inverse function f^{-1}(x) has the domain which is the range of the function f(x) (all possible values of f(x)), so the domain of inverse function is {-7,-1,8,14}

6 0
3 years ago
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