<span>you'll find the denominator is zero which isn't possible, so you'll get undefined. </span>
Answer:
B is the answer i think
Step-by-step explanation:
because its lenear and positive...
Hi there! the best way of solving this is picturing out what the graph might look like. Let's assume you had the graph of a parabola y=x^2. You know that for every x you substitute, there'd always be a value for y. Thus, the domain is ALL REAL NUMBERS or from -INFINITY to + INFINITY. The range on the other hand is different. We know that any number raised to the second power will always yield a positive integer or 0. Thus, y=x^2 won't have any negative y-values as the graph opens upward. Therefore, the range is: ALL REAL NUMBERS GREATER THAN OR EQUAL TO 0. or simply: 0 to +INFINITY.
<span>On the other hand, a cubic function y=x^3 is quite different from the parabola. For any x that we plug in to the function, we'd always get a value for y, thus there are no restrictions. And the domain is ALL REAL NUMBERS or from -INFINITY to + INFINITY. For the y-values, the case would be quite similar but different to that of the y=x^2. Since a negative number raised to the third power gives us negative values, then the graph would cover positive and negative values for y. Thus, the range is ALL REAL NUMBERS or from -INFINITY to + INFINITY. Good luck!!!:D</span>
Answer:
(16, 6 )
Step-by-step explanation:
Basically you are expressing the points (dots ) in coordinate form
Locate x = 16 on the x- axis ( between 14 and 18 )
Then go vertically up to the point at (16, 6 )
the solutions to a general quadratic equation is
X=-b±√b²-4ac/2a ,, When ax²+bx+c=0
the discriminant is the expression under the radical b²−4ac
Part A:
discriminant is (-16)² - 4(9)(60) = -1904
there are two complex solutions
Part B:
4x² + 8x − 5 = 0
4x² + 10x -2x - 5=0
2x ( 2x + 5) - 1(2 x + 5) =0
(2x + 5)(2x-1) = 0
/ \
/ \
2x+5 = 0 2x-1 =0
x = -5/2 x = 1/2
