All categories

3 years ago

What is the constant of variation for xy=12

Mathematics

1 answer:

IrinaVladis [17]3 years ago

6 0

Xy=k where k is constant of variation
12 is the constant of variation

You might be interested in

Ebony's bank balance 1st reached $400 on Day 4. The last day her balance was $400 was Day 8.

Tanya [424]

Ebony’s bank balance is $400 was Day 8 and this can be explained below.

<h3>How to explain the information?</h3>

Jade cannot be true because for every transaction made during withdraws, a little amount of money is collected and as such the amount will never remain $400.

Note that the scenario for the above to occur is:

On day 4, her balance = $400

On day 5, her balance = $40

On day 6, her balance = $400

On day 7, her balance = $400

On day 8, her balance = $400

Another fact is that since she deposited 400 at first, on day 8, there will be a 0 increase so the amount will still be 400.

Therefore, if the above is the case, one can deduce the fact that her balance will still be $400.

Learn more about estimation on:

brainly.com/question/107747

#SPJ1

6 0

1 year ago

I don't know how to solve this

Assoli18 [71]

4x? because you solve the exponent

8 0

3 years ago

Multiply (3x-4)(6x+7

Hitman42 [59]

Multiplying ,you get 18x^2+21x-24x-28. You simplify to get:

18x^2-3x-28.

8 0

3 years ago

In the triangle shown we can find the angle θ as follows calculator

Xelga [282]

Given a right triangle with hypothenus of measure 34, the side opposite the angle θ of measure 30, and the side adjacent the angle theta of measure 16.

$\sin\theta= \frac{opposite}{hypothenuse} = \frac{30}{34} = \frac{15}{17} \\ \\ \Rightarrow \theta=\sin^{-1}\left( \frac{15}{17} \right) \\ \\ \\ \cos\theta= \frac{adjacent}{hypothenuse} = \frac{16}{34} = \frac{8}{17} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{8}{17} \right) \\ \\ \\ \tan\theta= \frac{opposite}{adjacent} = \frac{30}{16} = \frac{15}{8} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{15}{8} \right)$

3 0

3 years ago

What is the oblique asymptote of the function f(x) x^2+x-2/x+1

Mamont248 [21]

$f(x)=\displaystyle\frac{x^2+x-2}{x+1}=\frac{x(x+1)-2}{x+1}=x-\frac{2}{x+1}
$

$\displaystyle\lim_{x\to\infty}\left\{f(x)-x \right\}=\lim_{x\to\infty}\frac{2}{x+1}=\lim_{x\to\infty}\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}
=0$

Consequently, t<span>he limit of $f(x)$ as x approaches infinity is $x$ .

In other words, $f(x)$ approaches the line y=x,
</span><span>
so oblique asymptote is y=x.

I'm Japanese, if you find some mistakes in my English, please let me know.</span>

3 0

3 years ago