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svp [43]
3 years ago
12

Please help! Explain to me how you found your answer.

Mathematics
2 answers:
GenaCL600 [577]3 years ago
7 0

Yo sup??

To solve this problem u just need the basics of trigonometry

So

Sinx=opposite side/hypotenuse

Cosx=adjacent side/hypotenuse

Tanx=sinx/cosx=opposite/adjacent

Now just plug in the values

Sinx=2√3/4

=√3/2

Cosx=2/4

=1/2

Tanx=√3

If u had a better knowledge of trigonometry then u will find that x is 60° or π/3

Hope this helps

Katena32 [7]3 years ago
6 0

Step-by-step explanation:

Sin: 2 square root 3: 4

Cos: 2:4

Tan: 2 square root 3: 2

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IgorLugansk [536]
The formula for the area of a circle is A = (pi)r^2
So if you plugged in the numbers you would get this equation A = (3.14)97^2
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6 0
3 years ago
11.4÷35.7 answer estimated<br>(15 points)(brainliest included)<br>(incorrect answer=report)​
Gre4nikov [31]

Answer:

0.31932773109

Step-by-step explanation:

  • Set it up
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8 0
3 years ago
Jayden buys a game for ​$10.77 and some lotion for ​$3.12. He pays with a ​$20 bill. Estimate how much change he should​ get, to
guapka [62]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What did they do wrong?
Tatiana [17]
They didn’t distribute the -1 (which is the “-”) correctly into the numbers in the parentheses. After distributing correctly, step 1 should be
8a+3+5a-2a

Hope this helps :)
5 0
3 years ago
Write in standard form 2+4i over 1+i
Levart [38]
So hmm a conjugate, is pretty much just, the same binomial, but with a different sign in the middle, so, a + b, has a conjugate of a - b
or -a + b, has a conjugate of - a  - b, or c - d, has a conjugate of c +d, and so on

anyway, the idea being, to "rationalize" the expression, namely, getting rid of the pesky radical in the denominator

so, we'll multiply the expression by 1, since anything times 1 is just itself

however, bear in mind, that 1, can be a/a, or b/b, or cheese/cheese, or anything/anything

so, we'll multiply the top and bottom of the fraction, by the conjugate of the denominator

anyhow, that said   \bf \cfrac{2+4i}{1+i}\cdot \cfrac{1-i}{1-i}\implies \cfrac{(2+4i)(1-i)}{(1+i)(1-i)}\implies \cfrac{2-2i+4i-4i^2}{1-i^2}\\\\&#10;-----------------------------\\\\&#10;\textit{recall that }i^2=-1\\\\&#10;-----------------------------\\\\&#10; \cfrac{2-2i+4i-4(-1)}{1-(-1)}\implies \cfrac{2+2i+4}{2}\implies \cfrac{6+2i}{2}\implies \cfrac{6}{2}+\cfrac{2i}{2}&#10;\\\\\\&#10;\boxed{3+i}
7 0
3 years ago
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