Hi there! You have to remember these 6 basic Trigonometric Ratios which are:
- sine (sin) = opposite/hypotenuse
- cosine (cos) = adjacent/hypotenuse
- tangent (tan) = opposite/adjacent
- cosecant (cosec/csc) = hypotenuse/opposite
- secant (sec) = hypotenuse/adjacent
- cotangent (cot) = adjacent/opposite
- cosecant is the reciprocal of sine
- secant is the reciprocal of cosine
- cotangent is the reciprocal of tangent
Back to the question. Assuming that the question asks you to find the cosine, sine, cosecant and secant of angle theta.
What we have now are:
- Trigonometric Ratio
- Adjacent = 12
- Opposite = 10
Looks like we are missing the hypotenuse. Do you remember the Pythagorean Theorem? Recall it!
Define that c-term is the hypotenuse. a-term and b-term can be defined as adjacent or opposite
Since we know the value of adjacent and opposite, we can use the formula to find the hypotenuse.
- 10²+12² = c²
- 100+144 = c²
- 244 = c²
Thus, the hypotenuse is:

Now that we know all lengths of the triangle, we can find the ratio. Recall Trigonometric Ratio above! Therefore, the answers are:
- cosine (cosθ) = adjacent/hypotenuse = 12/(2√61) = 6/√61 = <u>(6√61) / 61</u>
- sine (sinθ) = opposite/hypotenuse = 10/(2√61) = 5/√61 = <u>(5√61) / 61</u>
- cosecant (cscθ) is reciprocal of sine (sinθ). Hence, cscθ = (2√61/10) = <u>√61/5</u>
- secant (secθ) is reciprocal of cosine (cosθ). Hence, secθ = (2√61)/12 = <u>√</u><u>61</u><u>/</u><u>6</u>
Questions can be asked through comment.
Furthermore, we can use Trigonometric Identity to find the hypotenuse instead of Pythagorean Theorem.
Hope this helps, and Happy Learning! :)
A. 12*x + 2*y = 90
6*x + 2*y = 60
To write system of equation you just need to sum all solo and ensemble acts and make it equal to how much they will last for.
B. Here we can multiply second equation with negative 1 and add it to first equation.
6x = 30
x = 5
now we replace x in any of 2 equations to get y.
30+2y = 60
2y = 30
y=15
Solo acts are 5 minutes long and ensemble acts are 15 minutes long
Answer:
![\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}a+e&b+f\\c+g&d+h\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2Be%26b%2Bf%5C%5Cc%2Bg%26d%2Bh%5Cend%7Barray%7D%5Cright%5D)
We have:
![A=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]\\and\\B=\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D)
And we need to express as a single matrix:
![A+B=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6+(-2)&-3+8\\10+3&-1+(-12)\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6-2&5\\13&-1-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=A%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%2B%28-2%29%26-3%2B8%5C%5C10%2B3%26-1%2B%28-12%29%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6-2%265%5C%5C13%26-1-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
The answer is:
![\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
It is expressed as a single matrix.
Answer:
a. 2
b. 6
c. 1
Step-by-step explanation:
The degree is the highest exponent on the variable in an expression
a. 2
b. 6
Both x and y have exponents of 3. To determine the degree, add the exponents together. 3+3=6
c. 1
When no exponent is present on the variable, it is always 1.