Every even number is 2 away from the last.
0, 2, 4, 6, 8, 10, 12, 14...etc.
If we had an even number p, then the next three even numbers would be
p+2, p+4, and p+6.
<em>(If we had an odd number p, then the next three even numbers would be</em>
<em>p+1, p+3, and p+5. I'm not sure if p is even is implied in the question. Technically the answer would be p - p mod 2 + 2, where p is an interger...that gets into more technical function stuff, though.)</em>
4 cos² x - 3 = 0
4 cos² x = 3
cos² x = 3/4
cos x = ±(√3)/2
Fixing the squared cosine doesn't discriminate among quadrants. There's one in every quadrant
cos x = ± cos(π/6)
Let's do plus first. In general, cos x = cos a has solutions x = ±a + 2πk integer k
cos x = cos(π/6)
x = ±π/6 + 2πk
Minus next.
cos x = -cos(π/6)
cos x = cos(π - π/6)
cos x = cos(5π/6)
x = ±5π/6 + 2πk
We'll write all our solutions as
x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k
Answer:
With the vertex
We see that b = -10 and a = 2 and then the vertes wuld be:
And the best option is:
b. (-10,2)
Step-by-step explanation:
For this problem we have the following function:
And if we compare this expression with the general expression for a parabola given by:
With the vertex
We see that b = -10 and a = 2 and then the vertes wuld be:
And the best option is:
b. (-10,2)
Answer:
x=y−6
Step-by-step explanation:
Answer:
3 is not a function because you cant have the same input with different outputs
4 is a function because each input has one output
Step-by-step explanation: