Answer:
2 real solutions
Step-by-step explanation:
Remember this messy thing?
The <em>quadratic formula</em>, as it's called, gives us the roots to any quadratic equation in standard form (ax² + bx + c = 0). The information on the <em>type</em> of roots is contained entirely in that bit under the square root symbol (b² - 4ac), called the <em>discriminant</em>. If it's non-negative, we'll have <em>real</em> roots, if it's negative, we'll have <em>complex roots</em>.
For our equation, we have a discrimant of (-3)² - 4(6)(-4) = 9 + 96 = 105, which is non-negative, so we'll have real solutions, and since quadratics are degree 2, we'll have exactly 2 real solutions.
To find the slope, we must use the formula stated below.
Slope = Change in Y / Change in X = (Y1 - Y2) / (X1 - X2)
Since our coordinates are given as (X, Y), we must pick a coordinate to be #1, and the other coordinate will be #2.
I'll use the first coordinate as our #1.
(3 - 6) / (2 - 7) = (-3) / (-5) = 3/5 which is our answer
Answer:
Step-by-step explanation:
Answer:
The 4th graph
Step-by-step explanation:
To determine which graph corresponds to the f(x) = \sqrt{x} we will start with inserting some values for x and see what y values we will obtain and then compare it with graphs.
f(1) = \sqrt{1} = 1\\f(2) = \sqrt{2} \approx 1.41\\f(4) = \sqrt{4} = 2\\f(9) = \sqrt{9} = 3
So, we can see that the pairs (1, 1), (2, 1.41), (4, 2), (3, 9) correspond to the fourth graph.
Do not be confused with the third graph - you can see that on the third graph there are also negative y values, which cannot be the case with the f(x) =\sqrt{x}, the range of that function is [0, \infty>, so there are only positive y values for f(x) = \sqrt{x}
Answer:
C
Step-by-step explanation:
