Answer:
x = a/(a² + b²) or x = -1/a
Step-by-step explanation:
a(a²+ b²)x² + b²x - a =0
Use the quadratic equation formula:

1. Evaluate the discriminant D
D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²
2. Solve for x


Answer:
A
Step-by-step explanation:
From the graph, we can see that the <em>temperature increased initially</em>, then <em>started to decrease</em>. <u>All of these NOT in a constant rate, though</u>.
Roughly, the temperature was increasing from start till around 5.5 hours. Then on the temperature decreased.
So we can eliminate B, C, and D. That leaves us with correct answer as A.
Answer:
0.67 cm/s
Step-by-step explanation:
The area of a square is given by :
....(1)
Where
x is the side of a square

Differentiating equation (1) wrt t.

When A = 81cm², the side of the square, x = 9 cm
Put all the values,

So, the length of the side of a square is changing at the rate of 0.67 cm/s.
The relation is <em>not a function</em> if any x-value is repeated.
Assuming you have
(x, y) = (3, 10), (__, 20)
if x = 3, the relation is not a function.
7/8-5/8= 2/8 and if you reduce 2/8 by 2 it becomes 1/4. So 1/4 is your answer. Hope This Helps :D