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4 years ago

Graph the ellipse with equation x squared divided by nine plus y squared divided by sixteen = 1

Mathematics

2 answers:

Yakvenalex [24]4 years ago

8 0

Here is the picture of the answer above

gladu [14]4 years ago

6 0

X²/9 + y²/16 = 1

The general for mula of an ellipse is:

(x-h)²/a² + (y-k)²/b² = 1. Since h = k = 0, this ellipse
x²/9 + y²/16 = 1
passes by the center O. (Note that a = 3 and b = 4)

Moreover sin b>a so the majore axis is (on the y-axis) and a the minor axis (on the x-axis) [Its shape is as a vertical egg].

On a system of perpendicular axis that intercept in O., take on the x-axis 2 points A & B with respective coordinate A(3.0) and B(-3,0)
On the y-axis your report A'(0,+4) and B'(0,-4). So you have the vertex of the ellipse, then it's easy to draw it

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3 years ago

If p is a positive integer,then p(p+1)(p-1) is always divisible by?

ElenaW [278]

I am not quite sure what the choices are, but the answer to that problem is:

If p is a positive integer, then p(p+1)(p-1) is always divisible by “an even number”.

The explanation to this is that whatever number you input to that equation, the answer will always be an even number. This is due to the expression p(p+1)(p-1) which always result in a even product.

For example if p=3, then (p+1)(p-1) becomes (4)(2) giving you a even number.

And if for example if p=2, then (p+1)(p-1) becomes (3)(1) which gives an odd product, but we still have to multiply this with p therefore 2*3 = 6 which is even product. The outcome is always even number.

<span>Answer: From the choices, select the even number</span>

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3 years ago

7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

$_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\$

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

$_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\$

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

$_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\$

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>

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2 years ago