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FromTheMoon [43]
4 years ago
5

Which equation justifies why ten to the one third power equals the cube root of ten?

Mathematics
2 answers:
oee [108]4 years ago
8 0

For this case we must find the justification of:

10 ^ {\frac {1} {3}} = \sqrt [3] {10}

By definition of properties of powers and roots we have to meet:

a ^ {\frac {m} {n}} = \sqrt [n] {a ^ m}

So, if we have:

\sqrt [3] {10} = \ \sqrt [3] {10 ^ 1} = 10 ^ {\frac {1} {3}}

Other property states that:

(a ^ n) ^ m = a ^ {n * m}

So, the expression: "Ten to the one third power all raised to the third power" is represented as:

(10 ^{\frac {1} {3}}) ^ 3 = 10 ^ {\frac {3*1} {3}} = 10 ^ 1 = 10

ANswer;

Option B

Annette [7]4 years ago
4 0

Answer: The answer would be (10^1/3)³ = 10 ^(1/3×3) =10

Step-by-step explanation:

I took the test on FLVS

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Help mee quick and explain to me in an easy way as posible
yawa3891 [41]

Bags of limes = 4

Total limes in each bag = m

Total limes = 4m

ATQ

4m - 7(4)= 16

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Must click thanks and mark brainliest

3 0
3 years ago
Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)
viktelen [127]
\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}

The first line segment can be parameterized by \mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle with 0\le t\le1. Denote this first segment by C_1. Then

\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt
=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt
=108

The second line segment (C_2) can be described by \mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle, again with 0\le t\le1. Then

\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt
=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt
=-\dfrac{229}3

Finally,

\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3
5 0
4 years ago
Which expression gives the unit length of the line segment with the endpoints (-12, 8) and (2, 8)
Mashcka [7]
They have the same y coordinate, 8, so all we need to figure out is the distance between -12 and 2 of this horizontal line.
to do this add 12 and 2, you should get 14.
now using this information, find the expression that is equal to 14.
|-12|+|2|=14
|-12|-|2|=10
|8|+|8|=16
|8|-|8|=0
as we can see, the first expression would be the correct one.
5 0
3 years ago
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Verdich [7]

So we want to write a fraction that is less than 5/6 and has a denominator of 8. So lets start from 1/1 that is greater then 5/6 and doesn't have 8 as a denominator. Lets take one half from 1/1 and well get 1/2.

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Bagels: six in a bag Apples: eight in a bag Cookies: twelve in a box Juice Boxes: nine in a box Find the least number of package
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72/4 kids is 18 lunches
7 0
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