Answer:
yes
Step-by-step explanation:
2,4,6,8,10,12 so on
If Each side of an equilateral triangle<span> is 10 m. ... Thus </span>triangle<span> APC is a right</span>triangle<span>. The length of CA is 10 m, and the length of PC is 5 m, and hence you can use Pythagoras' theorem to find the length of AP, which </span>is the height<span> of the </span>triangle<span>ABC.
</span>If Each side of an equilateral triangle<span> is 10 m. ... Thus </span>triangle<span> APC is a right</span>triangle<span>. The length of CA is 10 m, and the length of PC is 5 m, and hence you can use Pythagoras' theorem to find the length of AP, which </span>is the height<span> of the </span>triangleABC.
Ok so find 1's
remember that 4/4=1 also x/x=1 those are ones
also remember
if you had
10/30 that equals 10/10 times 1/3=1 times 1/3
so find ones
4/10=2/2 times 2/5=1 times 2/5=2/5
Answer:
Final answer is
.
Step-by-step explanation:
Given problem is
.
Now we need to simplify this problem.
![\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)
![\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)
![\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)
Hence final answer is
.