All categories

3 years ago

Joe took $12 to the arcade each game cost $.50 to play. this situation can be modeled with the equation y=-.50x + 12.

Mathematics

1 answer:

inysia [295]3 years ago

7 0

Answer:

Step-by-step explanation:

The slope is what you multiply x by in a linear equation

You might be interested in

Question state in photo

vovangra [49]

Answer:

I believe it is B

Step-by-step explanation:

3 out of 4 options to land on are less than 5

5 0

3 years ago

Read 2 more answers

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

1/2 + 5/3 - 1<br>------------------<br> 3/4

Sav [38]

Hey there :)

$\frac{ \frac{1}{2}+ \frac{5}{3} -1 }{ \frac{3}{4} }$
↓ Is the same as
$\frac{1}{2} + \frac{5}{3} -1$ ÷ $\frac{3}{4}$

Since it is a division by fraction, we can multiply by what is called the reciprocal
( The attached picture might help you )
( $\frac{1}{2} + \frac{5}{3} - 1$ ) × $\frac{4}{3}$
↓ For this part, we need the common denominator, which is 6
$\frac{1(3)}{2(3)}+ \frac{5(2)}{32)} \frac{1(6)}{1(6)}$ × $\frac{4}{3}$

$\frac{3}{6}+ \frac{10}{6} - \frac{6}{6}$ × $\frac{4}{3}$
$\frac{7}{6}$ × $\frac{4}{3}$
$\frac{14}{9} = 1\frac{5}{9}$
↑
Your final answer

7 0

3 years ago

What is the solution please! i need this!!asap!!

Vlad1618 [11]

I believe the answer is no solutions

7 0

3 years ago

Give an example of two monomials with a quotient of -3n²/m

natima [27]

Hmmmm I’ll have to calculate this

3 0

3 years ago