Ask question

Ask question

All categories

2 years ago

8

What is the square or square root of 6 willing to give 24 points

1 answer:

evablogger [386]2 years ago

6 0

There is no square root of 6 all you have to do is set up a square root box and put the 6 in it then you split it into factors next you put them in another square root box finally multiply, then you get your simplest form.

You might be interested in

Hello, I would like help with this, please.

disa [49]

Answer:

slope= 1/2

c

Step-by-step explanation:

part a)

m= 4 - 0 / 8 - 0

= 1/2

5 0

2 years ago

You just got a free ticket for a boat ride, and you can bring along 4 friends! Unfortunately, you have 6 friends who want to com

Sliva [168]

Answer:

Step-by-step explanation:you can take 5

5 0

2 years ago

Read 2 more answers

Help please (will give brainliest)

klemol [59]

I belive you would subtract 18 from 90????

3 0

3 years ago

Read 2 more answers

A. y = -1.33x; 2.67<br> B. y = -0.75x; 1.50<br> C. y = 1.33x; -2.67<br> D. y = 0.13x; -0.25

Crazy boy [7]

Answer:

c is the correct answer for this problem

Step-by-step explanation:

7 0

2 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

2 years ago