The table shows the relationship between r and s, where s is the independent variable.

On July 9, Mifflin Company receives a $10,200, 90-day, 6% note from customer Payton Summers as payment on account. What entry sh

Ludmilka [50]

Answer:

Since on July 9, Mifflin Company receives a $ 10,200, 90-day, 6% note from customer Payton Summers as payment on account, to determine what entry should be made on July 9 to record receipt of the note the following calculation must be performed :

90 days = 3 months

6/12 x 3 = 1.5%

10,200 x 1,015 = 10,353

Therefore, a debt cancellation for $ 10,200 must be made in the company's accounting records, plus an interest generation for $ 153, which will be justified by the cash income of $ 10,353.

4 0

2 years ago

How to know if a function is periodic without graphing it ?

zhenek [66]

A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

$\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2$

and

$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

$\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}$

The first two equations are satisfied whenever $k\in\{0,\pm4,\pm8,\ldots\}$ , or more generally, when $k=4n$ and $n\in\mathbb Z$ (i.e. any multiple of 4).

The second two are satisfied whenever $k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}$ , and more generally when $k=\dfrac{10n}7$ with $n\in\mathbb Z$ (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

$\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2$

$\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5$

More generally, it can be shown that

$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

is periodic with period $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

4 0

2 years ago

The heaviest freshwater fish caught in region A weighs 209 lb, and the heaviest freshwater fish caught in region B weighs 629 l

Colt1911 [192]

Answer:

$Region\ A= \frac{209}{2240}\\\\Region\ B= \frac{629}{2240}$

Step-by-step explanation:

The English ton, also known as the long ton, is equivalent to 2,240 pounds. Therefore, the fractional parts, of each of the two fish, of the English ton are:

Fish caught in region A:

$A= \frac{209}{2240}$

Fish caught in region B:

$B= \frac{629}{2240}$

Both fractions are irreducible.

4 0

2 years ago

Which of the following are roots of the polynomial function? Check all that apply.

lutik1710 [3]

You know 1 is not a root because the sum of the coeffcients of the equation is 14, not zero.

It is fairly easy to try 3 by synthetic division (see attachment), which tells you that 3 is a root and the remaining quadratic factor is x²-3x-5. The quadratic formula tells you the roots of that factor are

... x = (-b±√(b²-4ac))/(2a) = (3±√29)/2

The appropriate choices are

... C. (3-√29)/2

... D. (3+√29)/2

... F. 3

_____

The quadratic formula tells you the solution to

... ax²+bx+c=0

is x = (-b±√(b²-4ac))/(2a)

We have a=1, b=-3, c=-5.

8 0

2 years ago

Is 1/3 closest to -1 or 0?

Nata [24]

Answer:

1/3 is closer to 0 than it is closer to -1

Step-by-step explanation:

<------------------(-1)--0---(1/3)----------------->

As you can see here, 1/3 is closer to 0 than -1

7 0

2 years ago

Read 2 more answers