Answer:
The system of equations is:
x + y = 15
xy = 15
Step-by-step explanation:
solve the first for y and substitute in the second:
y = 15-x
x(15-x) = 15
15x - x� = 15
15x - x� - 15 = 0
-x� + 15x - 15 = 0
x� - 15x + 15 = 0
Answer:
D. Minimum at (3, 7)
Step-by-step explanation:
We can add and subtract the square of half the x-coefficient:
y = x^2 -6x +(-6/2)^2 +16 -(-6/2)^2
y = (x -3)^2 +7 . . . . . simplify to vertex form
Comparing this to the vertex for for vertex (h, k) ...
y = (x -h)^2 +k
We find the vertex to be ...
(3, 7) . . . . vertex
The coefficient of x^2 is positive (+1), so the parabola opens upward and the vertex is a minimum.
Given:difference in the mean weight gain is 0.60 gramsstandard deviation of the difference in sample mean is 0.305
68% confidence interval for the population mean difference is a) 0.305
0.60 + 1 * 0.3050.60 + 0.305 = 0.9050.60 - 0.305 = 0.295
95% confidence interval for the population mean difference is c) 0.61
0.60 + 2 * 0.3050.60 + 0.61 = 1.210.60 - 0.61 = -0.01
Answer:
In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is (2,1).
Step-by-step explanation:
There is really no need to use any quadratics or roots.
( Consider the same problem on the plain number line first. )
How do you find the number between 2 and 5 which is twice as far from 2 as from 5?
You take their difference, which is 3. Now splitting this distance by ratio 2:1 means the first distance is two thirds, the second is one third, so we get
4=2+23(5−2)
It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result R, then
R=P+23(Q−P)
so in your case we get
R=(0,−1)+23(3,3)=(2,1)
Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning |t⋅a⃗ |=t|a⃗ | for any positive scalar t)
Edit: We'll try to divide a distance s into parts a and b such that a is twice as long as b. So it's a=2b and we get
s=a+b=2b+b=3b
⇔b=13s⇒a=23s
