Answer:
The length of the diagonal of the trunk is 56.356011 inches
Step-by-step explanation:
According to the given data we have the following:
height of the trunk= 26 inches
length of the trunk= 50 inches
According to the Pythagorean theorem, to calculate the length of the diagonal of the trunk we would have to calculate the following formula:
length of the diagonal of the trunk=√(height of the trunk∧2+length of the trunk∧2)
Therefore, length of the diagonal of the trunk=√(26∧2+50∧2)
length of the diagonal of the trunk=√3176
length of the diagonal of the trunk=56.356011
The length of the diagonal of the trunk is 56.356011 inches
You sometimes need to rewrite a factor because the lead number is not between 1 and 10 as it needs to be in scientific notation. For instance, if we had lead numbers of 2 and 5, as we do in the following problem.
2x10^3 * 5x10^9
Then we get the answer of 10x10^12. We need to adjust this by moving the decimal place and increasing the exponent by 1. The correct answer would be 1x10^13 instead.
Multiply both sides of the equation by 12. Move the variables to the left-hand side and change its sign. Move the constant to the left-hand side and change its sign. Collect like terms. Add the numbers. Divide both sides of the equation by 12.
1/2 x - 5= - 1/2 x + 19/4
6x -20= - 6x + 57
6x + 6x= 57 + 20
12x= 57 + 20
12x= 77
ANSWER
x = 77/12
Alternative Form .
x = 6 5/12, x= 6.416
Answer:
x = 180
Step-by-step explanation:
First, you need to know
1. Double-angle formula:
cos(2x) = 
2. Pythagorean identity:
Back to your problem, replacing the variable by the above:
By Double-angle formula
By Pythagorean identity
Given 
, we know -1 < sinx < 1, for every x ∈ R
Answer:
The answer is "![\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-2%5C%5C%20-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B7%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%7D)
".
Step-by-step explanation:
![\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D)
Solve the L.H.S part:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D)
After calculating the L.H.S part compare the value with R.H.S:
![\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D%20%5C%5C%5C%5C)
In equation (i) multiply by 3 and subtract by equation (iii):
put the value of c in equation (i):
In equation (ii) multiply by 3 then subtract by equation (iv):
put the value of d in equation (iv):
The final answer is "".
