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3 years ago

Step by step on how to get j/-2+7=-12

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

3 0

Answer:

Step-by-step explanation:

(j/-2) + 7 = -12

I move the 7 over by subtracting since its a positive 7

j/-2 = -19

now multiply each side by -2 to cancel out the denominator

j = 38

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40 percent of 112.5%

lord [1]

First 112% is 1.12 divide it by 100 to get .0112 then mutiply it by 40 to get .448 so .448 is the answer. Hope this helps.

7 0

3 years ago

Find the least common denominator for these two rational expressions. B/b^2 -64 -7b/b^2+7b-8

san4es73 [151]

Answer:

The least common denominator is (b-8)(b+8)(b-1)

Step-by-step explanation:

We are given expression as

$\frac{b}{b^2-64}-\frac{7b}{b^2+7b-8}$

Firstly, we will factor both denominators

$b^2-64=b^2-8^2=(b-8)(b+8)$

$b^2+7b-8=(b+8)(b-1)$

so, we can plug it back

$\frac{b}{(b-8)(b+8)}-\frac{7b}{(b+8)(b-1)}$

First term denominator is

(b-8)(b+8)

Second term denominator is

(b+8)(b-1)

So,

Least common denominator will be

(b-8)(b+8)(b-1)

So, we get

$LCD=(x-8)(x+8)(x-1)$

3 0

3 years ago

Ons: Practice

Sedaia [141]

Answer:

$x = 18$

Step-by-step explanation:

Given

$\frac{x}{6} - 7 = -4$

Required

Solve for x

$\frac{x}{6} - 7 = -4$

Add 7 to both sides

$\frac{x}{6} - 7+7 = -4+7$

$\frac{x}{6} = -4+7$

$\frac{x}{6} = 3$

Multiply through by 6

$6 * \frac{x}{6} = 3 * 6$

$x = 3 * 6$

$x = 18$

7 0

3 years ago

What is a equation that goes through (-4,7) and (-2,1) points

s2008m [1.1K]

Answer:

wdym

Step-by-step explanation:

6 0

3 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

3 years ago