Answer:
We conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Step-by-step explanation:
Given the expression
![\sqrt[3]{200k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
so the expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
first solving
![\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
Apply radical rule: ![\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
![\sqrt[3]{k^{15}}=k^{\frac{15}{3}}=k^5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D%3Dk%5E%7B%5Cfrac%7B15%7D%7B3%7D%7D%3Dk%5E5)
then solving
![\sqrt[3]{200}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200%7D)
prime factorization: 200: 2³ · 5²
![=\sqrt[3]{2^3\cdot \:5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%5Ccdot%20%5C%3A5%5E2%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
![=\sqrt[3]{2^3}\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%7D%5Csqrt%5B3%5D%7B5%5E2%7D)
Apply radical rule:
![\sqrt[n]{a^n}=a,\:\quad \:a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%2C%5C%3A%5Cquad%20%5C%3Aa%5Cge%200)
so
![=2\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D2%5Csqrt%5B3%5D%7B5%5E2%7D)
Thus, the main expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
![=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Therefore, we conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Given :
A holiday meal cost 12.50 a person plus a delivery fee of $30 at we cater.
The same meal cost $15 a person with no fee at Good Eats.
To Find :
When does we cater become the better deal.
Solution :
Let , x is number of order .
Cost at cater , C = 12.5x + 30 .
Cost at Good Eats , G = 15x .
We need to find :
G > C

Therefore, after 12th order cater will be more value for money.
Hence, this is the required solution.
Answer:
21
Step-by-step explanation:
Boys:
Q1: 78
Q3: 140
IQR: 140 - 78 = 62
Girls:
Q1: 97
Q3: 180
IQR: 180 - 97 = 83
Difference:
83 - 62
21
There are 6 less boys then girls
there are 6 more girls then boys
there are 15 boys compared to 21 girls
Answer:
a) SAS
b) Volume of DEF is 4 times the Volume of ABC
Step-by-step explanation:
a) since they have one same/congruent angle (angle B and angle E) and other two sides are in the same ratio, they are similar by SAS similarity theorem.
40/20 = 30/15 = 2
Therefore Δ ABC is similar to Δ DEF
b) Volume of a prism
= base area × height
Let height be 'h cm'
With Base ABC:
Volume = ½×20×15×h = 150h cm³
With Base DEF:
Volume = ½×40×30×h = 600h cm³
600h/150h = 4
Prism with base DEF has a greater volume; it's 4 times of the volume of the prism with base ABC