$First~exercise! \\ We~ have `there ~some~ triangles.(2) \\ 1)~In~ a ~triangle~,the~degree~of~all~angles~is~180! \\ We~have~2~angles~known,so: \\ 180=x+45+56 \\ 180=101+x \\ x=79 \\ \\ Second~triangle! \\ We~have~to~find~2~angles! \\ An ~elongated`~angle~have~180~degrees! \\ 180=79+50+y \\ y=51. \\ In~the~second~triangle~,we~have~to~find~the`third~angle. \\ We~have~a~right~angle,so--\ \textgreater \ 90~degrees! \\ 180=90+51+z \\ z=39! \\ We~found~all~the~angles!(We~are`happy~now!!!~:))) \\ Second~exercise! \\ With~the~Pythagoras' Theorem,we~can~find~the~diagonal~in~this~ \\ rectangle! \\ c ^{2} =10 ^{2} +14 ^{2} \\ c ^{2} =100+196 \\ c= \sqrt{296} \\ c=17,20...$

8 0

3 years ago

Can any one help me and work 2 or 3 of the problems for me to understand it

Alekssandra [29.7K]

To solve these problems you need to use the quadratic formula which is listed below.

$\frac{-b +/- \sqrt{b^{2} - 4ac } }{2a}$

So, if you take #2.

The $x^{2}$ has a 1 in front of it. The x has a -3 in front of it. The last number is also -3. So:

a = 1
b = -3
c = -3

Then you can plug all of these into the equation and simplify. Which gives you:

$\frac{3 -/+ \sqrt{21} }{2}$

4 0

3 years ago

If sin(a) = cos(2a), then what is the value of a?

Mariulka [41]

Answer:

-90 , 30

Step-by-step explanation:

sina = cos(2a)

= 1 - 2(sina)^2

2(sina)^2 + sina -- 1 = 0

( sina + 1 )( 2sina - 1 ) = 0

sina = -1 or sina = 1/2

=> a = -90 degrees or => a = 30 degrees

3 0

3 years ago