First differences are 2, 4, 8, 16, which is a geometric sequence. The parent function is not linear (constant first difference) or quadratic (first difference increases by the same amount from one to the next). When the first differences are a geometric sequence, the underlying sequence is a geometric (exponential) sequence.
1st blank: exponential
Translation up adds a constant to each of the f(x) values.
2nd blank: f(x)
3rd blank: increased by 5<span>
For the last blank, you're looking for an (x, f(x)) pair that is translated to (x, f(x)+5).
4th blank: </span><span>(2, 16)</span>
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
Answer:
it's 
Step-by-step explanation:
Answer:
8oz = .5lbs
128lb = 2,048oz
64lbs = 1,024oz
1 1/2lbs = 24oz
5lbs = 80oz
1lb = 16oz
Step-by-step explanation:
Answer:
Expected value of profit is ≅ $15.80
Step-by-step explanation:
from the Question,
Let the variable X represents the expected value of profit. X is called as random variable because picking a number from 000-999 digits is a Random process.
P(win) = 
So, P(lose) = 
Suppose that,
we really want to win this lottery. so we can go to the store and spend $1000 to buy all ticket (from 000 - 999). This would ensure your winning of $500 with one of the tickets (for a $499 profit), but the other 999 would be losers (for a $999 loss).
What would be your average winnings on a per-ticket basis?
∑
= 
Here,
Standard deviation of the expected winnings
∑
= ∑ 
Taking square root of the variance to get the standard deviation:
SD(x) = 
≅ $15.80
Hence
The expected value of profit is ≅ $15.80
