f(x) = x²<span> - 5x + 1
f(-3) = (-3)</span>² - 5(-3) + 1
<span>
f(-3) = 25
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Answer: f(-3) = 25 (Answer C)
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Answer:
Step-by-step explanation:
Given
Required
Determine the domain
First, we need to solve for p
Substitute 70 for t in
Collect Like Terms
Solve for p
This implies that Larry can not invite up to 6 people.
Hence, the domain can be represented as:
So for this question, we're asked to find the quadrant in which the angle of data lies and were given to conditions were given. Sign of data is less than zero, and we're given that tangent of data is also less than zero. Now I have an acronym to remember which Trig functions air positive in each quadrant. . And in the first quadrant we have that all the trig functions are positive. In the second quadrant, we have that sign and co seeking are positive. And the third quadrant we have tangent and co tangent are positive. And in the final quadrant, Fourth Quadrant we have co sign and seeking are positive. So our first condition says the sign of data is less than zero. Of course, that means it's negative, so it cannot be quadrant one or quadrant two. It can't be those because here in Quadrant one, we have that all the trick functions air positive and the second quadrant we have that sign. If data is a positive, so we're between Squadron three and quadrant four now. The second condition says the tangent of data is also less than zero now in Quadrant three. We have that tangent of data is positive, so it cannot be quadrant three, so our r final answer is quadrant four, where co sign and seek in are positive.
You first convert everything to minutes
3hrs= 60x3=180mins
1hr= 60x1=60mins
then you add 20 to 180
200
and you also add 48 to 60
104
200-104=
96
Then you convert to hrs and mins
96-60=36
1hr 36mins
M can be any positive real number.
Explanation:
From f(x) = √(mx) ; if x is posive m has to be positive; if x is negative m has to be negative; if x is cero m can have any value, and the range will always be 0 or positve
From g(x) = m √x; x can only be 0 or positive and the range will have the sign of m.
Given we concluded that the range of f(x) can only be 0 or positive, then me can only be 0 or positive.
