All expressions whose sum represents the same vector as (r + s) + t. is Option b,c,e
<h3>What are the
expressions that equals (r + s) + t. ?</h3>
Where
r= (2,3)
s= (5,-3)
t= (-8,6)
Generally, the equation for statement is mathematically given as
(7,0)+(-8,6) ,(2,3)+(-3,3) and (-6,9)+(5,-3) are equal to (r+s)+t
Therefore
x=(2,3)+(5,-3)
x=(7,0)
Now we can calculate (r+s)+t as
(7,0)+(-8,6)
(7-8,0+6)
(-1,6)
For (b)
x=(7,0)+(-8,6)
x=(-1,6)
in this scenario x expressions is equal to (r+s)+t
For (c)
x=(2,3)+(-3,3,)
x=(-1,6)
in this scenario x expressions is equal to (r+s)+t
For (e)

x=(-1,6)
in this scenario x expressions is equal to (r+s)+t
In conclusion, all expressions whose sum represents the same vector as (r + s) + t. is b,c,e
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Answer:
![\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }](https://tex.z-dn.net/?f=%5Cboxed%7B5%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%7D)
Step-by-step explanation:
![\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)
![\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%20%5Cimplies%20%2810%5E%5Cfrac%7B1%7D%7B3%7D%20%29%5E%5Cfrac%7B1%7D%7B2%7D%20%3D10%5E%5Cfrac%7B1%7D%7B6%7D%20%3D%5Csqrt%5B6%5D%7B10%7D)
![\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%3D%5Csqrt%5B6%5D%7B10%7D)
![\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Ctext%7BSolving%20%7D%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)

![\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%3D%5Csqrt%5B3%5D%7B2%5Ccdot%205%5E3%7D%3D5%20%20%5Csqrt%5B3%5D%7B2%7D)
Once
![\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%20%5Csqrt%5B6%5D%7B10%7D)
We have
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
We can proceed considering the common base of exponentials
![\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%3D%20%202%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ccdot%20%202%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%20%3D%202%5E%7B%5Cfrac%7B3%7D%7B6%7D%20%7D%20%3D%202%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%3D%5Csqrt%7B2%7D)
Therefore,
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%205%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
Notice this is a geometric progression since each number multiplied by some factor equals the next number in the sequence, in this case,

Then by applying the formula for sum to infinity of a geometric progression,
Answer:
19/3 or 6.33 continuing
Step-by-step explanation:
using pemdas, do the exponent first
6-2=4
then multiply by 4
4x4 is 16
add 3^2 to this, which = 9
so 16+9-6 is on top which equals 19, this is over 3
so 19/3