Answer: a²+b² = -99/2
Step-by-step explanation:
Since we are given two equations, this equations will be solved simultaneously to get a² and b²
a³ - 3ab² = 47 ... 1
b³ - 3a² b = 52... 2
From 1, a(a² - 3b²) = 47...3
From 2, b(b² - 3a²) = 52... 4
Adding 3 and 4, we have;
a²+b²-3b²-3a² = 99 (note that a and b will no longer be part of the equations as they have been factored out)
a²+b²-(3b²+3a²) = 99
(a²+b²) -3(b²+a²)= 99
Taking the difference we have
- 2(a²+b²) = 99
a²+b² = -99/2
Answer: c=5.4
The two roots are 3/5 and 9/5
Step-by-step explanation:
assume 5x2−12x c=0 is supposed to be 5x^2 - 12x + c = 0
p = (12 + sqrt(144-20c))/10
q = (12 - sqrt(144-20c))/10
p-3q=0,
1.2 + 0.1sqrt(144-20c) +
-3.6 + 0.3sqrt(144-20c) = 0
-2.4 + 0.4sqrt(144-20c) +2.4 = 2.4
sqrt(144-20c) = 2.4/0.4 = 6
144-20c=36
144-36=20c
c = 108/20 = 5.4
5x^2-12x+5.4=0
x = 3/5 or x = 9/5
Here are a few things you'll need to know for this question:
- Domain: <u>The list of x-values that are possible on a line.</u>
- Range: <u>The list of y-values that are possible on a line.</u>
- Interval Notation: <u>Shows the domain/range using the endpoints</u>. Brackets mean that the endpoint is included, parentheses mean the endpoint is excluded. Ex: (2,10]. 2 is excluded, 10 is included.
- Closed Circles: <u>The endpoint is included.</u>
- Open Circles: <u>The endpoint is excluded.</u>
So firstly, let's look at the domain. We see that there is a closed circle at x = -2 and an open circle at x = 5. Using what we know, <u>the interval notation of the domain is [-2,5).</u>
Next, let's look at the range. We see that there is a closed circle at y = -5 and an open circle at y = 2. Using what we know, <u>the interval notation of the range is [-5,2).</u>
Answer:
Look Below
Step-by-step explanation:
m<1 could be 30
m<2 could be 60
m<3 could be 30
The answer is D one third of AB.
