Find the area of a circle with a circumference of 37.68 units.

Sergeeva-Olga [200]

1. Two of the main identities used in trigonometry for right triangles (Triangles that have an angle of 90°) are: Sine (Sin) and Cosine (Cos).

Sin x°=Opposite/Hypotenuse

Cos x°=Adjacent/Hypotenuse

2. The inverse of Sinx° is Cosec x°, then:

Cosec x°=1/Sin x°

3. The inverse of Cos x° is Sec x°, then:

Sec x°=1/Cos x°

4. Keeping this on mind, you have:

Cosc x°=Hypotenuse/Opposite

Sec x°=Hypotenuse/Adjacent

5. Therefore, the correct answer is:

The third option: Cosec x°=Hypotenuse/Opposite.

7 0

2 years ago

Read 2 more answers

A pack of brownies cost $6.60 If the price per brownie stays the same, what would be

cupoosta [38]

Answer:

19.8

Step-by-step explanation:

3x6.6

7 0

2 years ago

Read 2 more answers

Solve for x: 2/3+1/3x=2x

velikii [3]

Answer:

x=2/5 or

x=0.4

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Two mechanics worked on a car. The first mechanic charged $65 per hour, and the second mechanic charged $60 per hour. The mechan

Bess [88]

Answer:

First mechanic: 12.8 hours & Second mechanic:12.8 hours

First mechanic made 832$ & Second mechanic made 768$

Hope this is right. If not I apologize and please let me know:)

Step-by-step explanation:

First you would do 65+60=125. You do this to find how much they charge together. Then you would take the amount of money they make and divide that from 125: 1600/125=12.8. Now that you have 12.8 you would plug that in for each mechanic: 65 times 12.8=832 and 60 times 12.8=768. Now you have the amount of money each mechanic made and the amount of hours they worked.

4 0

2 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago