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
Since 83-59=24, 83 is added by 28 so your answer would be 111.
Answer:converge at 
Step-by-step explanation:
Given
Improper Integral I is given as

integration of
is -
![I=\left [ -\frac{1}{x}\right ]^{\infty}_3](https://tex.z-dn.net/?f=I%3D%5Cleft%20%5B%20-%5Cfrac%7B1%7D%7Bx%7D%5Cright%20%5D%5E%7B%5Cinfty%7D_3)
substituting value
![I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B%5Cinfty%20%7D-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)
![I=-\left [ 0-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%200-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)

so the value of integral converges at 
Answer:
(-5,7) (-5,8) (-5,3)
Step-by-step explanation:
Dont know if this will help but here you go