A summer camp wants to hire counselors aides to fill its staffing needs at minimum cost

Answers

Answer 1
Answer:

Answer:

8 counselors and 12 aids

Step-by-step explanation:

minimum number of staff to run camp = 20

Ratio of counselors to aids to work together = 2:3

To get the multiply factor = 20 / (2 counselors + 3 aids) = 20 / 5 =4

minimum number of counselors needed = 4 x 2 = 8 counselors

minimum number of aids needed = 4 x 3 = 12 aids


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Please help answer this

I need help with math can someone help me I’ll show the question?

Answers

What’s your question

Helpppp asap pleasee

Answers

Answer:

I think 1 is the right answer

Step-by-step explanation:

Help me with this again please

Answers

The one on the bottom left.

5/3 x − 10 = 1/3 x
What is the value of x?

Answers

Answer:

X=1.66666667 hope this hepled

Step-by-step explanation:

by simplifying both sides of the equation, then isolating the variable.
X =7.5

Please help ASAP. Worth 50 points

Answers

Answer:

the anser is 9.5+1.5n and the second ansr is 48x+16

Answer:

a

Step-by-step explanation:

HELP PLEZ TRIGONOMETRY!

Answers

(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha

Solution:

Given that we have to simplify:

(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) ---- eqn 1

We know that,

sin^2 x = 1 - cos^2 x

Substitute the above identity in eqn 1

(2\left(1-\cos ^(2) \alpha\right)-1)/(\sin \alpha+\cos \alpha)

Simplify the above expression

(2-2 \cos ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)

(1-2 \cos ^(2) \alpha)/(\sin \alpha+\cos \alpha) ------- eqn 2

By the trignometric identity,

(sin x + cos x)(sin x - cos x) = 1-2cos^2 x

Substitute the above identity in eqn 2

((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)

Cancel the common factors in numerator and denominator

((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)=\sin \alpha-\cos \alpha

Thus the simplified expression is:

(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha