Question

What is the square root of 7 over fourth root of 7?

Answer 1

\[\sqrt{7}/\sqrt[4]{7}\]

Answer 2

\[7^\frac{ 1 }{4 }\]

Answer 3

\(\Huge\color{blue}{ \sf \frac{\sqrt{7} }{\sqrt[4]{7}} }\)

Answer 4

When you say

\(\Huge\color{blue}{ \sf \sqrt{7} }\) it is same as saying \(\Huge\color{blue}{ \sf \sqrt[2]{7} }\)

Answer 5

\(\bf \large {\cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}=a^{\frac{{\color{blue} n}}{{\color{red} m}}}
\\ \quad \\ \quad \\

\cfrac{\sqrt{7}}{\sqrt[4]{7}}\implies \cfrac{\sqrt{7}}{1}\cdot \cfrac{1}{\sqrt[4]{7}}\implies 7^{\frac{1}{2}}\cdot \cfrac{1}{7^{\frac{1}{4}}}\implies 7^{\frac{1}{2}}\cdot 7^{-\frac{1}{4}} }\)

Answer 6

So, \(\Huge\color{blue}{ \sf \frac{\sqrt{7} }{\sqrt[4]{7}} }\) would be same as

\(\Huge\color{blue}{ \sf \frac{\sqrt[2]{7} }{\sqrt[4]{7}} =\frac{7^{1/2}}{7^{1/4} }=7^{1/4}}\)

Answer 7

and

\(\Huge\color{blue}{ \sf 7^{1/4}= \sqrt[4]{7^1}=\sqrt[4]{7}}\)

Answer 8

Thank you so much! I had just figured it out like a minute ago but ts good to know that I got it right! thank you all so much! Would anyone be willing to help me with another one?

Answer 9

You welcome :)

Answer 10

I can try but I stink at math.

Answer 11

ok great thanks! here it is
Which exponential function goes through the points (1, 16) and (4, 128)?

Answer 12

the pattern that I see is

(x,y) --> (x+1, y × 2)

Answer 13

how might that be seen as an exponential function that would go through that line? I have some choices if those might help

Answer 14

Post the choices please.

Answer 15

here ya go and sorry I should've posted them with the question

f(x) = 4(4)x

f(x) = 8(2)x

f(x) = 8(2)-x

f(x) = 4(4)-x

Answer 16

f(x) = 4(4)^x

f(x) = 8(2)^x

f(x) = 8(2)^-x

f(x) = 4(4)^-x

Answer 17

Yeah that's better.

Answer 18

is it b?

Answer 19

lets look at the first one.
f(x) = 4(4)^x

16= 4(4)^1
64= 4(4)^2
256= 4(4)^3

you can see A doesn't work . Lets check choice B.

f(x) = 8(2)^x

16 = 8(2)^1
32 = 8(2)^2
64= 8(2)^3
128 = 8(2)^4

yep it works

Answer 20

ok thank you soooo much!!

Answer 21

You are welcome !