10 the second power is 10². 10²=10*10=100. 100*18.72=1,872. Hoped I helped. :)
Answer:
11
Step-by-step explanation:
1.80m + 0.6s = 12.00
[substitute the value of m into the equation]
1.80×(3) + 0.6s = 12
5.4 + 0.6s = 12
[make s the subject of the formula]
0.6s = 12 - 5.4
0.6s = 6.6
[divide both sides of the equation by 0.6]
0.6s / 0.6 = 6.6 / 0.6
s = 11
To prove that 11 is correct
[substitute the value of m and s into the equation]
1.80m + 0.6m = 12.00
1.80×(3) + 0.6×(11) = 12.00
5.4 + 6.6 = 12.00
12.00 = 12.00
Answer:
<h2>a = 11</h2>
Step-by-step explanation:
![\dfrac{5+2\sqrt3}{7+4\sqrt3}\qquad\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{5+2\sqrt3}{7+4\sqrt3}\cdot\dfrac{7-4\sqrt3}{7-4\sqrt3}=\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{7^2-(4\sqrt3)^2}\qquad\text{use FOIL}\\\\=\dfrac{(5)(7)+(5)(-4\sqrt3)+(2\sqrt3)(7)+(2\sqrt3)(-4\sqrt3)}{49-4^2(\sqrt3)^2}\\\\=\dfrac{35-20\sqrt3+14\sqrt3-8(3)}{49-(16)(3)}=\dfrac{(35-24)+(-20\sqrt3+14\sqrt3)}{49-48}\\\\=\dfrac{11-6\sqrt3}{1}=11-6\sqrt3](https://tex.z-dn.net/?f=%5Cdfrac%7B5%2B2%5Csqrt3%7D%7B7%2B4%5Csqrt3%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%2Bb%29%28a-b%29%3Da%5E2-b%5E2%5C%5C%5C%5C%3D%5Cdfrac%7B5%2B2%5Csqrt3%7D%7B7%2B4%5Csqrt3%7D%5Ccdot%5Cdfrac%7B7-4%5Csqrt3%7D%7B7-4%5Csqrt3%7D%3D%5Cdfrac%7B%285%2B2%5Csqrt3%29%287-4%5Csqrt3%29%7D%7B7%5E2-%284%5Csqrt3%29%5E2%7D%5Cqquad%5Ctext%7Buse%20FOIL%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%285%29%287%29%2B%285%29%28-4%5Csqrt3%29%2B%282%5Csqrt3%29%287%29%2B%282%5Csqrt3%29%28-4%5Csqrt3%29%7D%7B49-4%5E2%28%5Csqrt3%29%5E2%7D%5C%5C%5C%5C%3D%5Cdfrac%7B35-20%5Csqrt3%2B14%5Csqrt3-8%283%29%7D%7B49-%2816%29%283%29%7D%3D%5Cdfrac%7B%2835-24%29%2B%28-20%5Csqrt3%2B14%5Csqrt3%29%7D%7B49-48%7D%5C%5C%5C%5C%3D%5Cdfrac%7B11-6%5Csqrt3%7D%7B1%7D%3D11-6%5Csqrt3)
Your question seems a bit incomplete, but for starters you can write
![\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}](https://tex.z-dn.net/?f=%5Cdfrac1%7B1-%5Csin%20x%7D-%5Cdfrac%7B%5Csin%20x%7D%7B1%2B%5Csin%20x%7D%3D%5Cdfrac%7B1%2B%5Csin%20x%7D%7B%281-%5Csin%20x%29%281%2B%5Csin%20x%29%7D-%5Cdfrac%7B%5Csin%20x%281-%5Csin%20x%29%7D%7B%281%2B%5Csin%20x%29%281-%5Csin%20x%29%7D%3D%5Cdfrac%7B1%2B%5Csin%20x-%5Csin%20x%281-%5Csin%20x%29%7D%7B%281-%5Csin%20x%29%281%2B%5Csin%20x%29%7D)
Expanding where necessary, recalling that
![(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x](https://tex.z-dn.net/?f=%281-%5Csin%20x%29%281%2B%5Csin%20x%29%3D1-%5Csin%5E2x%3D%5Ccos%5E2x)
, you have
![\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%2B%5Csin%20x-%5Csin%20x%281-%5Csin%20x%29%7D%7B%281-%5Csin%20x%29%281%2B%5Csin%20x%29%7D%3D%5Cdfrac%7B1%2B%5Csin%20x-%5Csin%20x%2B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%3D%5Cdfrac%7B1%2B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D)
and you can stop there, or continue to rewrite in terms of the reciprocal functions,
![\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x](https://tex.z-dn.net/?f=%5Cdfrac%7B1%2B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%3D%5Csec%5E2x%2B%5Ctan%5E2x)
Now, since
![1+\tan^2x=\sec^2x](https://tex.z-dn.net/?f=1%2B%5Ctan%5E2x%3D%5Csec%5E2x)
, the final form could also take
![\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1](https://tex.z-dn.net/?f=%5Csec%5E2x%2B%5Ctan%5E2x%3D%5Csec%5E2x%2B%28%5Csec%5E2x-1%29%3D2%5Csec%5E2x-1)
or