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N76 [4]
2 years ago
12

RESOLVER POR TRIGONOMETRIC RATIOS

Mathematics
2 answers:
VikaD [51]2 years ago
8 0

I'm assuming that we're looking for the angle of that symbol.

{cos}^{ - 1} =  \frac{adj}{hyp}  =  \frac{4.4}{11}

= 66.42182152 \: degrees

round as you wish.

Rashid [163]2 years ago
6 0

Answer:

Can you please show more context.

Step-by-step explanation:

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If ³√a=am , what is m​
Zina [86]

\sqrt[3]{a}  =  {a}^{m}  \\  =  >  { a}^{ \frac{1}{3} }  =  {a}^{m}  \\  =  > m =  \frac{1}{3}

<u>Answer</u><u>:</u>

<u>m =  \frac{1}{3}</u>

Hope you could understand.

If you have any query, feel free to ask.

8 0
2 years ago
What is the value of the missing number in the sequence<br><br> 4,12,36,__,324,972,2916
stiv31 [10]
108 
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12 times 3 =36 
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4 0
3 years ago
A person invests 7000 dollars in a bank. The bank pays 5.5% interest compounded
kotykmax [81]

The time required to get a total amount of $13,200.00 with compounded interest on a principal of $7,000.00 at an interest rate of 5.5% per year and compounded 12 times per year is 11.559 years. (about 11 years 7 months)

Answer:

t = 11.559 years

<h3>Compound Interest </h3>

Given Data

  • Principal P= $7000
  • Rate r= 5.5%
  • Final Amount A = $13200
  • Time t = ??

(about 11 years 7 months)

Calculation Steps:

First, convert R as a percent to r as a decimal

r = R/100

r = 5.5/100

r = 0.055 per year,

Then, solve the equation for t

t = ln(A/P) / n[ln(1 + r/n)]

t = ln(13,200.00/7,000.00) / ( 12 × [ln(1 + 0.055/12)] )

t = ln(13,200.00/7,000.00) / ( 12 × [ln(1 + 0.0045833333333333)] )

t = 11.559 years

Learn more about compound interest here:

brainly.com/question/24924853

6 0
2 years ago
Translate the sentence into an inequality. Twice the difference of a number and 6 is at most – 29. Use the variable w for the un
____ [38]

Answer:2w-6<-29

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
A and B are supplementary angles. If A= (4x – 30)° and m
AfilCa [17]

Answer:

82 degrees

Step-by-step explanation:

Hope this helps! ^^

3 0
3 years ago
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