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2 years ago

6

DUE RIGHT NOW!!!!! PLEASE HELP!!!!!

1 answer:

elixir [45]2 years ago

6 0

Gfurknegwuwn nakwubwkw. Jahsken

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If (x + 2) is a factor of x2 - 6x2 + kx + 10, k =

Natali5045456 [20]

Answer:

10.7

Step-by-step explanation:

add multiple than subtract what you multiple and add a decimal

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Use FTC2 to find the derivative

labwork [276]

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2 years ago

Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

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3 years ago

What is the factor of 31

Novosadov [1.4K]

Answer:

1 and 31

Step-by-step explanation:

They both divide by it's self to get the same number.

6 0

3 years ago

Read 2 more answers

Need help on number 2 plz I appreciate it thank you so much

vladimir1956 [14]

Answer:

$\angle PQR$ , $\angle SQR$ , and $\angle PQS$

Or

angle PQR, angle SQR and angle PQS

Step-by-step explanation:

The three different angles in the diagram are angle PQR, angle SQR and angle PQS.

Another way of writing this is using an angle sign before the alphabets follows. Thus:

$\angle PQR$ , $\angle SQR$ , and $\angle PQS$

6 0

2 years ago