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Yuri [45]
3 years ago
13

Please answer asap. pretty sure I know the answer just need to be certain.

Mathematics
1 answer:
astra-53 [7]3 years ago
4 0
  • Answer:

<em>p = 4$</em>

  • Step-by-step explanation:

<em>p + 25%p = $5</em>

<em>p + 25p/100 = 5</em>

<em>⁴⁾p + p/4 = ⁴⁾5</em>

<em>4p + p = 20</em>

<em>5p = 20</em>

<em>p = 20 : 5</em>

<em>p = 4$</em>

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- 6/39 simplified. :)​
stepan [7]

2/13_

Step-by-step explanation:

6/39_=

_6÷39/3

=2/39_

=11÷2/39_

=2/13_

4 0
1 year ago
NEED HELP ON THIS QUESTION PLEASE
Ilia_Sergeevich [38]

Answer:

Sample space= 1, 2, 3, 4, 5, 6

Outcomes= 2, 4, 6

Probability= Number of outcomes/ total number of outcomes

= 3/6 times 2( as 2 dices are rolled)

=6/12

=1/2

Therefore, the probability of getting 2 even number is 1/2

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%28sinx%5E%7B2%7D%20theta%29%5Cfrac%7Bx%7D%7By%7D%281%2Bcostheata%29" id="TexFormula1" title="
beks73 [17]

The result of expanding the trigonometry expression \sin^2(\theta) * (1 + \cos(\theta)) is cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

<h3>How to evaluate the expression?</h3>

The expression is given as:

\sin^2(\theta) * (1 + \cos(\theta))

Express \sin^2(\theta) as 1 - \cos^2(\theta).

So, we have:

\sin^2(\theta) * (1 + \cos(\theta)) =  (1- \cos^2(\theta)) * (1 + \cos(\theta))

Open the bracket

\sin^2(\theta) * (1 + \cos(\theta)) =  1 + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Express 1 as cos°(Ф)

\sin^2(\theta) * (1 + \cos(\theta)) =  cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Hence, the result of expanding the trigonometry expression \sin^2(\theta) * (1 + \cos(\theta)) is cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Read more about trigonometry expressions at:

brainly.com/question/8120556

#SPJ1

3 0
2 years ago
A school has 2 computer labs. Each lab has 30 computers. A total of
Korolek [52]

Answer:

(2*30)-6

Step-by-step explanation:

8 0
3 years ago
Help me please help me
ziro4ka [17]
Yes it’s a function
7 0
2 years ago
Read 2 more answers
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