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2 years ago

Given the diagram below, what is cos(45")? 45° 6 A. √2 B. 6/√2 C. 1/√2 D. 3√2

Mathematics

1 answer:

Gnesinka [82]2 years ago

3 0

C. 1/root2

Based on the information, the hypotenuse will be 6root2 because it is a 45, 45, 90 triangle. The two legs will be 6. Using SOH CAH TOA, we know that for cosine, it will be adjacent/hypotenuse. The adjacent side is 6 and the hypotenuse is 6root2. The result will be (6)/(6root2). After simplifying, we get
1/root2.

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If f(x) = 2x-12, evaluate -2f(3) <br><br> PLEASE HELP!!!!!!

mel-nik [20]

I think that’s it f(x)=2x−12

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2 years ago

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Third-degree, with zeros of −3 , − 2 , and 1 , and passes through the point ( 3 , 11 ) .

Masja [62]

Answer:

y = ( x + 3 ) * ( x + 2 ) * ( x - 1 )* 11/60

Step-by-step explanation:

Lets y = f(x) in a cartesian coordinates

Having three zeros:

x = -3 ⇒ x = -2 ⇒ and x = 1

That meas that if x takes the above mentioned values " y " must be zero

therefore y must be of the form

f (x) = y = ( x + 3 ) * ( x + 2 ) * ( x - 1 ) (1)

In that case for y to be zero one of the factors should be zero or

y = 0 x + 3 = 0 and x = - 3 is a zero of the function y .

The same reasoning applies for the other two roots

Now we have to evaluate the other condition.

According to problem statement the function passes through the point ( 3, 11 ) , that means that when x = 3 , y have to be 11, therefore we plug in equation (1) that value to see what happens

y = ( x + 3 ) * ( x + 2 ) * ( x - 1 )

11 = ( 3 + 3 ) * ( 3 + 2 ) + ( 3 - 1) = 6*5*2 = 60

Then we adjust the expression (1) to meet the condition of the function passing through point ( 3 , 11) as:

y = ( 3 + 3 ) * ( 3 + 2 ) + ( 3 - 1) * 11/60 (2)

and check to see if we did right

for y to be zero x can be x = - 3 x = -2 and x = 1 in all these cases y = 0. And if x = 3 in equation (2) y = 11. And that what we want to shown. Then the solution is:

y = ( x + 3 ) * ( x + 2 ) * ( x - 1 )* 11/60

8 0

3 years ago

Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you

adell [148]

Answer:

1) The solution of the system is

$\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) The solution of the system is

$\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

Step-by-step explanation:

1) To solve the system of equations

$\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right$

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

$\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 2: Swap rows 1 and 2

$\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 3: $\left(R_1=\frac{R_1}{6}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 5: $\left(R_2=\frac{R_2}{3}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 6: $\left(R_3=R_3+R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]$

Step 7: $\left(R_3=\left(\frac{6}{37}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 8: $\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 9: $\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 10: Rewrite the system using the row reduced matrix:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) To solve the system of equations

$\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right$

using the row reduction method you must:

Step 1:

$\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 2: $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 3: $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 5: $\left(R_2=\left(2\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 6: $\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 7: $\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]$

Step 8: $\left(R_3=\left(- \frac{2}{117}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 9: $\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 10: $\left(R_2=R_2-\left(15\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 11:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

8 0

3 years ago

How many equal sections divide a directed line segment if it is to be partitioned with a 2:7 ratio? 2, the least number given in

Elodia [21]

9514 1404 393

Answer:

9, the sum of the numbers given in the ratio

Step-by-step explanation:

The reduced ratio 2 : 7 means that two parts correspond to one length and 7 parts correspond to the other length. That is, there are a total of 2+7 = 9 parts in the full length of the segment.

It is convenient to think of the line segment as being divided into 9 equal parts.

_____

<em>Additional comments</em>

It is often helpful to consider the total number of "ratio units" in a given ratio. For example, in this problem, the shorter segment is 2/9 of the whole, and the longer one is 7/9 of the whole.

It can also be useful to consider what a "ratio unit" represents. If the whole segment is 27 inches long, then each of the 9 ratio units will represent 3 inches, for example. This means the division 2:7 is 2(3 in):7(3 in) = 6 in : 21 in.

3 0

3 years ago

If the point (x,square root 3/2) is on the unit circle, what. is x?

galina1969 [7]

Answer:

$x=\pm\frac{1}{2}$

Step-by-step explanation:

On a unit circle, $(x,y)=(\cos\theta,\sin\theta)$ , so $\sin\theta=\frac{\sqrt{3}}{2}$ in this case. If you look at the attached circle, the only time that the y-coordinate is $\frac{\sqrt{3}}{2}$ is when $x=-\frac{1}{2}$ and $x=\frac{1}{2}$ , which correspond to angles of $\theta=\frac{2\pi}{3}$ and $\theta=\frac{\pi}{3}$ respectively

3 0

3 years ago