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Naily [24]
3 years ago
9

Find the trigonometric ratios for sin C, cos C, tan C.

Mathematics
1 answer:
bonufazy [111]3 years ago
5 0
SinC=9/41
cosC=40/41
tanC=41/9
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Together Louisa and kill scored more than 30 points in the basketball game. If Kill scored 12 points how many points P did Louis
Romashka-Z-Leto [24]
Attached the solution with work shown.

8 0
3 years ago
Sophie and Simon are peeling a pile of potatoes for lunch in the cafeteria. Sophie can peel all the potatoes by herself in 45 mi
tensa zangetsu [6.8K]
Short Answer: 18 minutes
Remark

The answer to this problem is less than the smallest time of the two people working together. that fact lets out C and D (38 minutes and 75 minutes). Now you have to choose 15 minutes and 18 minutes. There's a method. No guessing needed.

Givens
Let the time for Sophie = S
Let the time for Simon = M
Let the job to completion = 1 
S = 45 minutes
M = 30 minutes

Step One
Convert minutes to hours.
45 minutes = 45 / 60 = 3/4 hour = 0.75 hour
30 minutes = 30 / 60 = 1/2 hour = 0.50 hour

Step Two
Set up the Equation
The formula is a form of job / hour.
Let the time = t that they both have to work

job = 1 in these problems.
1/S + 1/M = 1/t
1/0.75 + 1/0.5 = 1/t

Solve
1 ÷ 0.75 = 1.33333
1 ÷ 0.5 = 2

1.3333 + 2 = 3.33333 
3.3333 = 1 / t                    Multlply both sides by t
3.3333*t = 1
t = 1 / 3.333333333
t = 0.3 of an hour


1 hour = 60 minutes
0.3 hours = x              Cross Multiply
x = 60 * 0.3
x = 18 minutes

Answer working together it took them 18 minutes  <<<<< 
6 0
3 years ago
Read 2 more answers
A stock sale of 100 is called
Mars2501 [29]
A is da answer I thank
6 0
3 years ago
1- 4/5 divided by 2/3
Softa [21]
I hope this helps you

7 0
3 years ago
What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?
joja [24]

Answer:

23rd term of the arithmetic sequence is 118.

Step-by-step explanation:

In this question we have been given first term a1 = 8 and 9th term a9 = 48

we have to find the 23rd term of this arithmetic sequence.

Since in an arithmetic sequence

T_{n}=a+(n-1)d

here a = first term

n = number of term

d = common difference

since 9th term a9 = 48

48 = 8 + (9-1)d

8d = 48 - 8 = 40

d = 40/8 = 5

Now T_{23}= a + (n-1)d

= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118

Therefore 23rd term of the sequence is 118.

4 0
3 years ago
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