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finlep [7]
3 years ago
6

Solve the following triangle

Mathematics
2 answers:
MArishka [77]3 years ago
4 0

Answer:

69

Step-by-step explanation:

as , 77+x = 146(external angle) , so x = 146 - 77

so x = 69

Cloud [144]3 years ago
3 0

Answer:

Step-by-step explanation:

hope you'll now how to do it by ur own.

:)

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Can you help me figure this out please
Anna007 [38]

Answer:

C & D  

Step-by-step explanation:

4 0
3 years ago
Helpppppppppppppppp ill mark you brainlist write in y=mx+b form
Alex Ar [27]

Answer:

I think: 0 = 3x + b or b = -3x + 0 or x = -b + 0/3 or 3 = -b + 0/x

Sorry if you get this wrong..

4 0
2 years ago
Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
Solve for x<br> Urgent need help
fiasKO [112]
I believe you need to solve this using the quadratic formula!
To begin, this is what it is:
x= -b ± <span>√ b^2 - 4ac / 2a
Just plug in what you have in your problem...
2 being a, 13 being b, and -24 being c.
So we get:
x= -13 </span>± <span>√13^2 - 4(2)(-24) / 2(2)
x= -13 </span><span>± √169 - 8 (-24) / 4</span>
<span>x= -13 <span>± √169 + 192 / 4</span>
x= -13 </span>± √<span>361 / 4
The square root of 361 is 19.
So you have: -13 </span><span>± 19 / 4.
Here's where you take the equation </span>-13 <span>± 19 and put the addition and subtraction sign to use.
-13 - 19 = -32
and
-13 + 19 = 6
Now all is left to do is divide the two numbers by 4.
-32/4 = -8
and
6/4 = 3/2

x = -8, 3/2</span>
6 0
3 years ago
In Ellen's math class, there are 2 boys for every 3 girls . What is the the following ratio of boys to girls in the class ?
IrinaK [193]
Based on the ratio of boys/girls We can infer that the only possibility for Ellen's math class would be answer B. 14/21
4 0
3 years ago
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