Answer:
True, false, true, true.
Step-by-step explanation:
The roots zeros of a quadratic function are the same as the factors of the quadratic function. This is true because your roots are your factors—>(x-3) is a factor, x=3 is the root.
The roots zeros are the spots where the quadratic function intersects with the y-axis. No! Those are called y-intercepts!
The roots zeros are the spots where the quadratic function intersects with the x-axis. True. X-intercepts are your solutions. (x-3) graphed would the (3,0). That’s a solution.
There are not always two roots/zeros of a quadratic function, True. No solution would be when your quadratic doesn’t intersect the x-axis. One solution would be when your vertex would be on the x-axis. Two solutions is when your quadratic intersects the x-axis twice. Can there be infinite solutions? No. It’s either 0, 1, or 2 solutions.
Answer:
6
Step-by-step explanation:
The main idea of how to solve it is to isolate the variable. Whatever you multiply, divide, add, or subtract on one side, you must do the same to the other side so that both sides are equal.
3(x-1) = 21
3(x-1) / 3 = 21 / 3
x - 1 = 7
+ 1 + 1
x = 6
Answer:
4 i think
Step-by-step explanation:
<span>The name of the shape graphed by the function r ^ 2 = 9
cos (2 theta) is called the “<u>lemniscate</u>”. A lemniscate is a
plane curve with a feature shape which consists of two loops that meet at a
central point. The curve is also sometimes called as the lemniscate of
Bernoulli. </span>
Explanation:
The
period of coskθ is 2π/k. In this case, k = 2 therefore the
period is π.
r ^ 2 = 9 cos 2θ ≥0 → cos 2θ ≥0. So easily
one period can be chosen as θ ∈
[0, π] wherein cos 2θ ≥0.
As cos(2(−θ)) = cos2θ, the graph is symmetrical about the initial line.
Also,
as cos (2(pi-theta) = cos 2theta, the graph is symmetrical about the
vertical θ = π/2
A
Table for half period [0,π4/] is
adequate for the shape in Quarter1
Use symmetry for the other three quarters:
(r, θ) : (0,3)(3/√√2,π/8)(3√2/2,π/6)(0,π/4<span>)</span>
