Lines are perpendicular if the slope of one is the negative reciprocal of the other. In this case the slopes are -8/3 and 8/3, so while negative they are not reciprocal. So no, the two lines intersect but are not perpendicular.

4 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago

Is 56 a multiple of 6?

lbvjy [14]

No, 56 is a multiple of 7 though.

8 0

2 years ago

How do you use the product rule to find the derivative of y=x√⋅cos(x)y=sqrt(x)*cos(x) ?

Ipatiy [6.2K]

$y=\sqrt x\cos x\\
y'=(\sqrt x)'\cos x+\sqrt x (\cos x)'\\
y'=\frac{1}{2\sqrt x}\cos x+\sqrt x\cdot(-\sin x)\\
y'=\frac{1}{2\sqrt x}\cos x-\sqrt x\sin x\\
$

7 0

2 years ago