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vampirchik [111]
3 years ago
9

The value stored in variable s at the end of the execution of the loop could best be described as __________. Assume that all va

riables are of type int. z = 0; g = 0; s = 0; i = 0; while (i < 50) { scanf("%d", &t); s = s + t; if (t >= 0) g = g + 1; else z = z + 1; i = i + 1; }
Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Answer:

It will describe the sum of the numbers scanned.

As we can see whatever you scan or take input it will be added to s without any condition. So at the end, sum of all scanned number will be save in variable s

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What are the remaining trig functions? and how do i solve for them? pls help
erastovalidia [21]

You're told that tan(<em>θ</em>) is positive, but

tan(<em>θ</em>) = sin(<em>θ</em>)/cos(<em>θ</em>)

and you're also told that sec(<em>θ</em>) = 1/cos(<em>θ</em>) = -3. So if cos(<em>θ</em>) is negative, sin(<em>θ</em>) must also be negative. In turn, both sec(<em>θ</em>) = 1/cos(<em>θ</em>) and csc(<em>θ</em>) = 1/sin(<em>θ</em>) are also negative.

Now, recall the Pythagorean identity,

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Multiply through both sides by 1/cos²(<em>θ</em>) to get an alternate form of the identity,

1 + tan²(<em>θ</em>) = sec²(<em>θ</em>)

Solve for tan(<em>θ</em>) (which we know is positive):

tan(<em>θ</em>) = √(sec²(<em>θ</em>) - 1) = 2√2

Right away, we get

cot(<em>θ</em>) = 1/tan²(<em>θ</em>) = 1/(2√2) = √2/4

Since sec(<em>θ</em>) = -3, it follows that cos(<em>θ</em>) = -1/3.

Then

tan(<em>θ</em>) = sin(<em>θ</em>)/cos(<em>θ</em>)   ==>   sin(<em>θ</em>) = 2√2 × (-1/3) = -2√2/3

and so

csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -3/(2√2)

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3 years ago
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Solnce55 [7]
Is the answer A and C?
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4 years ago
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KIM [24]

Answer:

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Step-by-step explanation:

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3 years ago
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Answer:

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