Hang on ,it is already proven through given .
So The proof is of one line only
- ray FEH bisects <DFG (Given)
Hence proved .
20
U have to divide it by 3 the ratio is 1:3
Penjelasan langkah demi langkah:
1)
![= 243^{\frac{2}{3} }\\= (\sqrt[3]{243})^2\\= 7^2\\= 49](https://tex.z-dn.net/?f=%3D%20243%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%7D%5C%5C%3D%20%28%5Csqrt%5B3%5D%7B243%7D%29%5E2%5C%5C%3D%207%5E2%5C%5C%3D%2049)
2) √32 +3√18-2√50
= √16*2 +3√9*2-2√25*2
= 4√2 + 3(3√2)-2(5√2)
= 4√2 + 9√2-10√2
= 13√2-10√2
= 3√2
3) 1000 ⅔×64⅙
![= (\sqrt[3]{1000}) ^2 \times (2^6)^{1/6} \\= 10^2 \times 2\\= 100 \times 2\\= 200](https://tex.z-dn.net/?f=%3D%20%28%5Csqrt%5B3%5D%7B1000%7D%29%20%5E2%20%5Ctimes%20%282%5E6%29%5E%7B1%2F6%7D%20%20%5C%5C%3D%2010%5E2%20%5Ctimes%202%5C%5C%3D%20100%20%5Ctimes%202%5C%5C%3D%20200)
4) 3/4+√2

5) 2√3×√18
= 2√3×√9*2
= 2√3×3√2
= (2*3)(√3*√2)
= 6√6
6) 12/3+√3
= 4+√3
7) √1000—2√40
= 10 -2 (√4*10)
= 10-2(2√10)
= 10 - 4√10
8) 2- ¹+3-¹

9)

Jika pernyataannya opsional, penyebutnya adalah 1
10) 2√3×√18
= 2√3×√9*2
= 2√3×3√2
= (2*3)(√3*√2)
= 6√6
Answer:
- Multiples of 3 = 3, 9
- Multiples of 5 = 5, 10
Step-by-step explanation:
In a list of 3 to 99, the first multiple of 3 will be none other than 3 itself because when multiplied by 1, you get three.
The second number will be: 3 * 2 = 6
In a list from 5 to 100, the first multiple will be 5 with 5 being a multiple of itself when multiplied by 1.
Second number = 5 * 2 = 10