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3 years ago

Simplify each expression. Use positive exponents.

Mathematics

2 answers:

mario62 [17]3 years ago

4 0

Answer:

Step-by-step explanation:

(A) The given expression is:

$m^3n^{-6}p^0$

Simplifying this expression, we get

= $m^3n^{-6}(1)$

= $\frac{m^3}{n^6}$

which is the required simplified form.

(B) The given expression is:

$\frac{a^4b^{-3}}{ab^{-2} }$

Simplifying the above equation using the exponent rule, we have

= $\frac{a^3}{b}$

which is the required simplified form.

$(x^{-2}y^{-4}x^3)^{-2}$

Simplifying this expression, we get

= $(\frac{x}{y^4})^{-2}$

= $(\frac{y^4}{x})^2$

= $\frac{y^8}{x^2}$

which is the required simplified form.

Semmy [17]3 years ago

3 0

To simplify each expression, we need to group all the terms that have the same bases and flock them if necessary to one side. One rule that applies is that negative exponents goes to the other side in order to be positive.

1. m3n-6p0 ; p0 =1 , so, m3/n6
2. a^4b^-3/ab^-2 = a^4a-1 / b^-2b^3 = a^3/b
3. (x^-2y^-4x^3) -2 = (x^-2x^3 y^-4 )^-2 = (x/y^4)^-2 = y^16/x^2

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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 $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .