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3 years ago

HELp Help Help... fnljwbfjwopfowq

Mathematics

1 answer:

makkiz [27]3 years ago

8 0

Answer:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
x-(3*x^3+8*x^2+5*x-7)=0
Step by step solution :Step 1 :Equation at the end of step 1 : x-((((3•(x3))+23x2)+5x)-7) = 0 Step 2 :Equation at the end of step 2 : x - (((3x3 + 23x2) + 5x) - 7) = 0 Step 3 :Step 4 :Pulling out like terms :
4.1 Pull out like factors :
-3x3 - 8x2 - 4x + 7 =
-1 • (3x3 + 8x2 + 4x - 7)
Checking for a perfect cube :
4.2 3x3 + 8x2 + 4x - 7 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: 3x3 + 8x2 + 4x - 7
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 3x3 - 7
Group 2: 8x2 + 4x
Pull out from each group separately :
Group 1: (3x3 - 7) • (1)
Group 2: (2x + 1) • (4x)

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Read 2 more answers

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

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

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

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

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