Consider the vectors u <-4,7> and v= <11,-6>
1 answer:
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Answer:
(4, -25)
Step-by-step explanation:
One way to answer this is to put the equation into vertex form.
f(x)= x^2 -8x -9 . . . . . given
Add and subtract the square of half the x-coefficient:
f(x) = x^2 -8x +(-8/2)^2 -9 -(-8/2)^2
f(x) = x^2 -8x +16 -25
f(x) = (x -4)^2 -25
Comparing this to the vertex form of a quadratic:
f(x) = a(x -h)^2 +k
we find that (h, k) = (4, -25). This is the vertex.
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<em>Alternate solution</em>
The line of symmetry for ...
f(x) = ax^2 +bx +c
is given by x = -b/(2a). For your given quadratic that line is ...
x = -(-8)/(2(1)) = 4
Evaluating f(4) gives the y-coordinate:
f(4) = 4^2 -8·4 -9 = -25
The vertex is (x, y) = (4, -25).
Answer:
Step-by-step explanation:
We need at least two points to write the equation of a straight line.
The recursive formula that Elijah wrote is:
When we substitute n=0, we get:
The points (0,30) and (1,37) lies on this line.
The equation of this line is of the form:
where b =30 is the y-intercept and m=7 is the slope.
We plug in these values to get:
Note that the slope of the line is equal to the common difference of the Arithmetic Sequence.
You could also use the two points to find the slope: