The true equations are 24 over 2 squared equals 50 minus 6 squared minus 8 and 56 divided by 8 plus 3 squared equals 2 times 2 cubed.
The equation one is given as
39- 3(6) = 6(10/4)+4
After solving we get,
39 - 18 = 15 +4
21 = 19
Which is not a true equation.
2nd equation we have
24/(2^2)=50 -6^2-8
24/4 = 50 -36 -8
6 = 50 - 44
6 = 6
which is a true equation.
3rd equation we have
56/8+3^2= 2(2^3)
7 +9 = 2(8)
16 = 16
which is true equation.
4th equation, we have
8^2-5^2 = 4^2-7
16-10 = 16-7
6 = 9
Which is not a true equation.
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The anwser is for the question tht u are looking for is no other then the first anwser A!!!
use sin^2(x) =1/2 * (1-cos2x)
cos 2x - use def cosx from table
f(x)=sin^2(0)
f'(x)=2*sin(x)*cos(x)
therefore f'(x)=sin(2x)
f''(x)=2cos(2x)
f'''(x)=-4sin(2x)=-4*f'(x)
<span>putting it into maclaurin's series</span>
<span>We get,</span>
<span> 0+0+ 2x^2/2! + 0...</span>
Answer:
It is not linear
Step-by-step explanation:
If you plug that function into Desmos graphing Calculator, you can see there are two leparate lines and they are not straight.