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allochka39001 [22]
3 years ago
12

Algebra 2 Help!

Mathematics
2 answers:
m_a_m_a [10]3 years ago
6 0
Base on the question that ask to calculate the sine, cosine, and tangent of 5pi/3 radians and base on my further calculation, I would say that the answer would be <span>sin Θ = -sqrt 3/2; cos Θ = -1/2; tan Θ = -sqrt 3. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification </span>
Oksana_A [137]3 years ago
3 0

Answer:

sin(\frac{5\pi }{3})=-\frac{\sqrt{3}}{2} ; cos(\frac{5\pi }{3})=\frac{1}{2} ; tan(\frac{5\pi }{3})=-\sqrt{3}

Step-by-step explanation:

we know that

\pi \ radians=180\ degrees

so

\frac{5\pi }{3}\ radians=300\ degrees

The angle belong to the IV quadrant

therefore

The sine is negative

The cosine is positive

The tangent is negative

360\°-300\°=60\°

so

sin(\frac{5\pi }{3})=-sin(60\°)=-\frac{\sqrt{3}}{2}

cos(\frac{5\pi }{3})=cos(60\°)=\frac{1}{2}

tan(\frac{5\pi }{3})=-tan(60\°)=-\sqrt{3}

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On a bicycle trail, the city is painting arrows like the one shown below:
mote1985 [20]

Answer:

151.5 cm^2

Step-by-step explanation:

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area of a triangle is 1/2 base times hight --> 13x3/2 = 19.5cm^2

now lets doi the rectangle that is 12 by 11

Length times hight is the area is 132

add both areas 132 + 19.5

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3 years ago
One dozen students each drop a brass tack six times from a height of six inches onto a level hard surface. They record the numbe
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Experimental Probability = 2/3

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3 years ago
Read 2 more answers
Y^3-9y^2+y-9 I don't know how to solve this can you help
aleksley [76]
I'm guessing your problem is this:
y³ - 9y² + y - 9 = 0
right?

In solving this problem, I recommend doing this:
y³ - 9y² + y - 9 = 0
Factor out a y² from the first two numbers in the problem:
y²(y - 9) + (y - 9) = 0
Separate the parentheses which means y - 9 goes on one side. The y² added a one since it came from the + 1 in the middle of expression. When you're separating parentheses like this you just take the outside numbers and combine them together. Since + 1 came from the outside of the (y - 9) and y² also was sitting on the outside of (y - 9) combine them to make y² + 1. Like this:
(y² + 1)(y - 9) = 0
Now separate your two parentheses to two separate problems:
(y² + 1) = 0    and    (y - 9) = 0
Now you're y² + 1 will equal:
y² = -1
y = √-1 <-- This number doesn't exist so it will be an imaginary number (i). If you guys didn't learn that in your class I recommend just leaving it as i for that part.
Now solve y - 9 = 0:
y = 9 <-- Since we added nine to both sides to get this.

So you're final answer should be y = i and 9
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Answer Are SSS,ASA,AAS,SAS Not enough Information <br> PLEASE HELP ME PLEASE
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Answer:

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Step-by-step explanation:

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