1q+2=-6q+3+8q+4
1q+2=2q+7
<span>-2q -2q
</span>-q+2=7
<span> -2 -2
</span>-q=<span>5
-1 -1
q=-5
</span>
Answer:
c
Step-by-step explanation:
mass will never change only weight will as weight is dependent on gravitational field strength
<h3>2
Answers: Choice C and choice D</h3>
y = csc(x) and y = sec(x)
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Explanation:
The term "zeroes" in this case is the same as "roots" and "x intercepts". Any root is of the form (k, 0), where k is some real number. A root always occurs when y = 0.
Use GeoGebra, Desmos, or any graphing tool you prefer. If you graphed y = cos(x), you'll see that the curve crosses the x axis infinitely many times. Therefore, it has infinitely many roots. We can cross choice A off the list.
The same applies to...
- y = cot(x)
- y = sin(x)
- y = tan(x)
So we can rule out choices B, E and F.
Only choice C and D have graphs that do not have any x intercepts at all.
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If you're curious why csc doesn't have any roots, consider the fact that
csc(x) = 1/sin(x)
and ask yourself "when is that fraction equal to zero?". The answer is "never" because the numerator is always 1, and the denominator cannot be zero. If the denominator were zero, then we'd have a division by zero error. So that's why csc(x) can't ever be zero. The same applies to sec(x) as well.
sec(x) = 1/cos(x)
It's a value you should probably memorize:
You can derive it using some trigonometric identities, other known values of cosine, and properties of the cosine function. For example, using the double angle identity for cosine:
If 
, then
and you probably know that 
, so
When we take the square root, we should take the positive root because 
whenever 
:
