Answer:
LHS.= Sin 2x /( 1 + cos2x )
We have , sin 2x = 2 sinx•cosx
And. cos2x = 2cos^2 x - 1
i.e . 1+ cosx 2x = 2cos^2x
Putting the above results in the LHSwe get,
Sin2x/ ( 1+ cos2x ) =2 sinx•cosx/2cos^2x
=sinx / cosx
= Tanx
.•. sin2x/(1 + cos2x)= tanx
Step-by-step explanation:
Excercise 1:
No, this is not simplified fully. The full answer would be <span>−7dk+14d−21k</span>−7, not <span>14d - 9 - 21k - 7dk + 2. So, it is not equivalent.
</span><span>
Excercise 4:
</span><span>7dk (2 - 3 - 1) - 7
</span>
Just multiply 7dk into (2) (-3) and (-1)
you'd get:
<span><span>−<span>14dk</span></span>+</span>−<span>7 after simplifying it fully.
</span>
p = parameter
w = width
w+2.75
p = 2w+2(w+2.75) = <span>4w</span>+<span>5.5 = 30
Answer to the last one:
w = 6.125</span>
I believe it would be 44.75 sq yards
Answer:
Ues
Step-by-step explanation:
Ues papa math or demos and that should help
Answer:
Final answer is ![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
.
Step-by-step explanation:
Given problem is ![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
.
Now we need to simplify this problem.
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
Hence final answer is .
