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3 years ago

Use double angle formulas to find the exact value of sin2x,cos2x, and tan2x

1 answer:

5 0

Sin2x = 2sinxcosx;
cos2x = (cosx)^2 - (sinx)^2;
tan2x = (sin2x)/(cos2x);

cosx = 5/13 from formula (sinx)^2 + (cosx)^2 = 1;

=> sin2x = 120/169;
.................................

You might be interested in

Convert the given system of equations to matrix form

yuradex [85]

Answer:

The matrix form of the system of equations is $\left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right] \left[\begin{array}{c}x&y&w&z&u\end{array}\right] =\left[\begin{array}{c}5&4&3\end{array}\right]$

The reduced row echelon form is $\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

The vector form of the general solution for this system is $\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Step-by-step explanation:

Convert the given system of equations to matrix form

We have the following system of linear equations:

$x+y+w+z-3u=5\\x-y-2w+z+2u=4\\2x+w-z+u=3$

To arrange this system in matrix form (Ax = b), we need the coefficient matrix (A), the variable matrix (x), and the constant matrix (b).

$A= \left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right]$

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]$

$b=\left[\begin{array}{c}5&4&3\end{array}\right]$

Use row operations to put the augmented matrix in echelon form.

An augmented matrix for a system of equations is the matrix obtained by appending the columns of b to the right of those of A.

So for our system the augmented matrix is:

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\1&-1&-2&1&2&4\\2&0&1&-1&1&3\end{array}\right]$

To transform the augmented matrix to reduced row echelon form we need to follow this row operations:

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\2&0&1&-1&1&3\end{array}\right]$

add -2 times the 1st row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\0&-2&-1&-3&7&-7\end{array}\right]$

multiply the 2nd row by -1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&-2&-1&-3&7&-7\end{array}\right]$

add 2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&2&-3&2&-6\end{array}\right]$

multiply the 3rd row by 1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&1&-3/2&1&-3\end{array}\right]$

add -3/2 times the 3rd row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 3rd row to the 1st row

$\left[\begin{array}{ccccc|c}1&1&0&5/2&-4&8\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 2nd row to the 1st row

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

Find the solutions set and put in vector form.

Interpret the reduced row echelon form:

The reduced row echelon form of the augmented matrix is

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

which corresponds to the system:

$x+1/4\cdot z=3\\y+9/4\cdot z-4u=5\\w-3/2\cdot z+u=-3$

We can solve for z:

$z=\frac{2}{3}(u+w+3)$

and replace this value into the other two equations

$x+1/4 \cdot (\frac{2}{3}(u+w+3))=3\\x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}$

$y+9/4 \cdot (\frac{2}{3}(u+w+3))-4u=5\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}$

No equation of this system has a form zero = nonzero; Therefore, the system is consistent. The system has infinitely many solutions:

$x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}\\z=\frac{2u}{3}+\frac{2w}{3}+2$

where u and w are free variables.

We put all 5 variables into a column vector, in order, x,y,w,z,u

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=\left[\begin{array}{c}-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}&\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}&w&\frac{2u}{3}+\frac{2w}{3}+2&u\end{array}\right]$

Next we break it up into 3 vectors, the one with all u's, the one with all w's and the one with all constants:

$\left[\begin{array}{c}-\frac{u}{6}&\frac{5u}{2}&0&\frac{2u}{3}&u\end{array}\right]+\left[\begin{array}{c}-\frac{w}{6}&-\frac{3w}{2}&w&\frac{2w}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Next we factor u out of the first vector and w out of the second:

$u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

The vector form of the general solution is

$\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

7 0

4 years ago

I need help with this one asap please and thank you

inna [77]

Answer:

Step-by-step explanation:

(m² - 3m + 2) / (m² - m)

we see due to a little bit of experience with expressions and multiplications of expressions that

(m² - 3m + 2) = (m - 2)(m - 1)

(m² - m) = m(m - 1)

so,

(m - 2)(m - 1) / (m(m - 1)) = (m - 2) / m

so, that's not it.

(m² - 2m + 1) / (m - 1)

we see again

(m² - 2m + 1) = (m - 1)(m - 1)

so,

(m - 1)(m - 1) / (m - 1) = m - 1

so, that's not it.

(m² - m - 2) / (m² - 1)

we see again

(m² - m - 2) = (m - 2)(m + 1)

and

(m² - 1) = (m + 1)(m - 1)

so,

(m - 2)(m + 1) / ((m + 1)(m - 1)) = (m - 2) / (m - 1)

yes, that is the solution.

(2m² - 4m) / (2(m - 2))

2m(m - 2) / (2(m - 2)) = 2m/2 = m

no, that is not a solution.

8 0

2 years ago

Solve x to the second power = 144

vfiekz [6]

x ^ 2 = 144

x = 12, x = -12

Check:

12 * 12 = 144

-12 * -12 = 144

6 0

3 years ago

Read 2 more answers

A jewellery shop sells 240 necklaces in a month. 180 were sold via the shops website, the rest were sold in a high street shop.

Rom4ik [11]

3:4
180 / 6 = 30 / 10 = 3
240 / 6 = 40 / 10 = 4
3 = website sales
4 = store sales
website : store
=
3 : 4

4 0

4 years ago

Because she has limited shelf space, she can't put out all her copies of the CD at once. On Monday morning, she stocked the disp

fenix001 [56]

Now, it look like there is some information missing in the answer. The whole problem should look like this:

Alicia Keys's new album As I Am is climbing the charts, and the manager of Tip Top Tunes expects to sell a lot of copies. Because she has limited shelf space, she can't put out all her copies of the CD at once. On Monday morning, she stocked the display with 40 copies. By the end of the day, some of the copies had been sold. On Tuesday morning, she counted the number of copies left and then added that many more to the shelf. In other words, she doubled the number that was left in the display. At the end of the day, she discovered that she had sold the exact same number of copies as had been sold on Monday. On Wednesday morning, the manager decided to triple the number of copies that had been left in the case after Tuesday. Amazingly, she sold the same number of copies on Wednesday as she had on each of the first two days! But this time, at the end of the day the display case was empty. How many copies of the As I Am CD did she sell each day?

Answer:

She sold 24 copies of the cd each day.

Step-by-step explanation:

In order to solve this problem we must first set our variable up. In this case, since we need to know what the number of sold cd's per day is, that will just be our variable:

x= Number of copies sold.

So we can start setting our equation up. So we take the first part of the problem:

"On Monday morning, she stocked the display with 40 copies. By the end of the day, some of the copies had been sold."

This can be translated as:

40-x

where this expression represents the number of copies left on the shelf by the end of monday.

"On Tuesday morning, she counted the number of copies left and then added that many more to the shelf."

so we represent it like this:

(40-x)+(40-x)

"In other words, she doubled the number that was left in the display."

so the previous expression can be simplified like this:

2(40-x)

"At the end of the day, she discovered that she had sold the exact same number of copies as had been sold on Monday."

so the expression now turns to:

2(40-x)-x this is the number of copies left by the end of tuesday.

"On Wednesday morning, the manager decided to triple the number of copies that had been left in the case after Tuesday."

this translates to:

3[2(40-x)-x]

This is the number of copies on the shelf by the begining of Wednesday.

"Amazingly, she sold the same number of copies on Wednesday as she had on each of the first two days! But this time, at the end of the day the display case was empty."

this piece of information lets us finish writting our equation:

3[2(40-x)-x] -x = 0

since there were no copies left on the shelf, then the equation is equal to zero.

So now we proceed and solve the equation for x:

3[2(40-x)-x] -x = 0

We simplify it from the inside to the outside.

3[80-2x-x]-x=0

3[80-3x]-x = 0

we now distribute the 3 so we get:

240-9x-x=0

we combine like terms so we get:

240-10x=0

we move the 240 to the other side of the equation so we get:

-10x=-240

and divide both sides into -10 so we get:

x=24

so she sold 24 copies each day.

5 0

4 years ago