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tensa zangetsu [6.8K]
3 years ago
5

Find the standard form of the equation of the ellipse with the given characteristics.

Mathematics
1 answer:
kicyunya [14]3 years ago
8 0
So hmm if you notice the picture below

between 2 and 16, there are 14 units, so, the center is half-way in between, thus, is at y = 7, or -2, 7

because, the major axis is over the y-axis, then the "a" component, goes under the fraction with the "y" in the numerator

\bf \textit{ellipse, vertical major axis}\\\\

\cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}, {{ k}}\pm a)\\\\
-----------------------------\\\\

\begin{cases}
b=4\\
a=7\\
h=-2\\
k=7
\end{cases}\implies \cfrac{(x+2)^2}{4^2}+\cfrac{(y-7)^2}{7^2}\implies \cfrac{(x+2)^2}{16}+\cfrac{(y-7)^2}{49}

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PLEASE HELP this is a missing assignment i will mark brainliest
chubhunter [2.5K]

The polynomial a(x) = -18x² - 6x + 12 is the dividend of the polynomial division

The quotient q(x) is 0 and the remainder r(x) is -18x² - 6x + 12

<h3>How to divide the polynomial?</h3>

The polynomial functions are given as:

a(x) = -18x² - 6x + 12

b(x) = 3x³ + 9x - 1

The quotient equation is given as:

a(x)/b(x) = q(x) + r(x)/b(x)

Since the degree of the dividend a(x) is less than the degree of the divisor b(x), then it means that the value of the quotient q(x) is:

q(x) = 0

And the remainder r(x) is:

r(x) = a(x)

Substitute known values

r(x) = -18x² - 6x + 12

Hence, the quotient q(x) is 0 and the remainder r(x) is -18x² - 6x + 12

Read more about polynomial division at:

brainly.com/question/25289437

5 0
2 years ago
N is a positive integer
Murrr4er [49]

Part (1)

n is some positive integer. Let's say for now that n is even. So n = 2k, for some integer k

This means n-1 = 2k-1 is odd since subtracting 1 from an even number leads to an odd number.

Now multiply n with n-1 to get

n(n-1) = 2k(2k-1) = 2m

where m = k(2k-1) is an integer

The result 2m is even showing that n(n-1) is even

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

Let's say that n is odd this time. That means n = 2k+1 for some integer k

And also n-1 = 2k+1-1 = 2k showing n-1 is even

Now multiply n and n-1

n(n-1) = (2k+1)(2k) = 2k(2k+1) = 2m

where m = k(2k+1) is an integer

We've shown that n(n-1) is even here as well.

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

So overall, n(n-1) is even regardless if n is even or if n is odd.

Either n or n-1 will be even. If you multiply an even number with any number, the result will be even.

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

Part (2)

n is some positive integer

2n is always even since 2 is a factor of 2n

2n+1 is always odd because we're adding 1 to an even number. The sequence of integers goes even,odd,even,odd, etc and it does this forever.

-----------

Another way to see how 2n+1 is odd is to divide 2n+1 over 2 and you'll find that we get (2n+1)/2 = 2n/2+1/2 = n+0.5

The 0.5 at the end is not an integer, so there's no way that (2n+1)/2 is an integer; therefore 2n+1 is odd.

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balandron [24]

Answer:

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Step-by-step explanation:

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3 years ago
Jenifer needs to purchase a computer for her company price of the computer is c
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If you provide more information then I could try to help :)

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