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Evgen [1.6K]
2 years ago
8

100 POINTS AND BRAINLIST

Mathematics
2 answers:
Lesechka [4]2 years ago
8 0

Answer:

\sf tan(X) = 2.14

\sf sin(X)= 0.91

\sf cos(X)= 0.42

Solve for tan(X):

\sf tan(x) = \dfrac{opposite}{adjacent}

\sf tan(X) = \dfrac{30.8}{14.4}

\sf tan(X) = 2.14

Solve for sin(X):

\sf sin(x)= \dfrac{opposite}{hypotensue}

\sf sin(X)= \dfrac{30.8}{34}

\sf sin(X)= 0.91

Solve for cos(X):

\sf cos(x)= \dfrac{adjacent}{hypotensue}

\sf cos(X)= \dfrac{14.4}{34}

\sf cos(X)= 0.42

Alla [95]2 years ago
3 0
  • X=B

All ratios remains same

\\ \rm\hookrightarrow sinX=\dfrac{P}{H}=\dfrac{30.8}{34}=0.9

\\ \rm\hookrightarrow cosX=\dfrac{B}{H}=\dfrac{14.4}{34.4}=0.42

\\ \rm\hookrightarrow tanX=\dfrac{sinX}{cosX}=\dfrac{0.9}{0.42}=2.14

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fenix001 [56]
No photo?? what are the choices
6 0
3 years ago
Read 2 more answers
PartA: Explain why x=6makes 4x-5< 19.
NARA [144]

Answer:

It doesn't.

Step-by-step explanation:

If you plug in 6 in the x spots you will have:

<h3>4(6)-5 < 19</h3>

If you solve that, you will get:

<h3>19< 19</h3>

So considering that they are the same number, this doesn't make sense because it should be an equal sign.

Have a nice day!!! Please mark Brainliest!

3 0
3 years ago
In how many ways can you arrange MATHEMATICS if duplication of the arrangement are not allowed?​
sukhopar [10]

Answer:

4989600 ways

Step-by-step explanation:

From the question,

The word MATHEMATICS can be arranged in n!/(r₁!r₂!r₃!)

⇒ n!/(r₁!r₂!r₃!) ways

Where n = total number of letters, r₁ = number of times M appears r₂ = number of times A appears, r₃ = number of times T appears.

Given: n = 11, r₁ = 2, r₂ = 2, r₃ = 2

Substitute these value into the expression above

11!/(2!2!2!) = (39916800/8) ways

4989600 ways

Hence the number of ways MATHEMATICS can be arranged without duplicate is 4989600 ways

7 0
2 years ago
Cost of a comic book: 3.95<br> Markup: 20%
Ghella [55]
The cost would be $4.74. 3.95 times 0.2 is 0.79. Add that to 3.95 and get 4.47
4 0
3 years ago
1. In a table of random digits, Each digit is to occur with a probability of 0.1
anyanavicka [17]
(A) Just because every digit has an equal chance of appearing does not mean that all will be equally represented. (See "gambler's fallacy")

(B) The experimental procedure isn't exactly clear, so assuming a table of digits refers to a table of just one-digit numbers, each with 0.1 chance of appearing (which means you can think of the digits 0-9), you should expect any given digit to appear about 0.1 or 10% of the time.

So if a table consists of 1000 digits, one could expect 7 to appear in 10% of the table, or about 100 times.
8 0
3 years ago
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