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dalvyx [7]
3 years ago
11

here is points my children <3 enjoy them just answer the question for the points: regards - drip king.

Mathematics
2 answers:
allochka39001 [22]3 years ago
8 0

Answer:

thxs <3

Step-by-step explanation:

Arte-miy333 [17]3 years ago
6 0

Answer:

Thank you

Step-by-step explanation:

Kind person

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Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )
Zina [86]

Answer:

\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

8 0
3 years ago
What is the value of x in simplest radical form ?
STALIN [3.7K]
Use Pythagorean theorem
a^2 + b^2 = c^2
12^2 + 35^2 = c^2
144 + 1225 = c^2
1369 = c^2
Sqrt 1369 = c
37 = c
4 0
3 years ago
Read 2 more answers
Simplify <br> 2x x y x 3
nikitadnepr [17]
Which is it
2x times y times 3 = 6xy
or,
2xxyx3= 6x^3y

7 0
3 years ago
Please help me right away.
velikii [3]
I’m pretty sure it’s c , sorry if it’s wrong tho
4 0
3 years ago
Is the network in d) an Euler circuit? Can this network can be traversed?
beks73 [17]

Answer:

If a graph is an Euler Circuit that mean that it can be traversed and begins and has all even verticies.  This allows you to start and stop at the same verticie.

Step-by-step explanation:

4 0
3 years ago
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