9514 1404 393
Answer:
(x, y) = (8, 2)
Step-by-step explanation:
Using the geometric mean rules for this geometry, we have ...
h = √(4·32) = 8√2
and also ...
a = √(4·36) = 12
b = √(32·36) = 24√2
For the problem at hand, ...
h = 8√2 = x√y ⇒ x = 8, y = 2
6 1/3
6 10/3 = 9 1/3
9 1/3 divided by 6 1/3 = 1.47368421053
Answer:
The slope is 2/3
Step-by-step explanation:
Explanation step by step is attached below :)
This is not an arithmetic sequence.
It does not go up or down by the same amount each time.
It is a geometric sequence
it is multiplied by 5 each time
'Vertex form', having just googled it, is another name for completing the square, which I have done extensively at school. This involved converting:
f(x) = ax^2 + bx + c --> f(x) = a(x-b)^2 + c
In the completed-square form, (b, c) are the co-ordinates of the vertex (which is the maximum/minimum point). So for part a:
- Here is the original
f(x) = x^2 + 12x + 11
- Halve the x-coefficient (the number before x) and use it as -b in the vertex form described above. You must then calculate the square of this number and minus it at the end because when you multiply it out, this is the surplus you will make along with x^2 + 12x:
f(x) = (x + 6)^2 - 36 + 11
- Tidy this up by collecting the constant (just number) terms together:
f(x) = (x + 6)^2 - 25
- Using the form a(x - b)^2 + c, we can work out that b = -6 (because -b = 6) and c = -25. This gives us the vertex (-6, -25) which is a minimum point because the graph is a positive quadratic, giving us the characteristing 'U' shape which has a bottom. If it were a negative quadratic (denoted by a negative x^2 coefficient), the vertex will be a maximum point because it has an 'n' shape instead.
- To solve the equation from here, make the function equal to zero:
(x + 6)^2 - 25 = 0
- Then take 25 to the other side:
(x + 6)^2 = 25
- Next, square-root both sides:
x + 6 = <span>±5
- Rearrange to finish:
x = -6 </span><span>± 5 = -1 or -11
- Therefore, the roots (solutions) to the equation x^2 + 12x + 11 = 0 are x = -1 or -11
This method will work the same for the other equations up there too, so I will leave them for you to do.
I hope this helps
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