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Temka [501]
2 years ago
14

Why is cube root of 9 equal to 9 to the one third power question mark

Mathematics
2 answers:
alexira [117]2 years ago
7 0

Answer:

See below

Step-by-step explanation:

\sqrt[3]{9}=9^{\frac{1}{3}}

(\sqrt[3]{9})^3=(9^{\frac{1}{3}})^3

9=9^{\frac{1}{3}*3

9=9^\frac{3}{3}

9=9^1

9=9

Remember that (a^b)^c=a^{b*c}

kozerog [31]2 years ago
3 0

Answer:

3

just because

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A glass cylinder is completely filled with 24x^5y^8 cubic inches of salt solution. The jug has a base area of 4x^5 square inches
xz_007 [3.2K]
We know that

volume of a cylinder=area of the base*height
area of the base=4x^5  in²
height=<span>24x^5y^8/ 4x^5 in
volume=</span>24x^5y^8 in³

height=24x^5y^8/ 4x^5 ------> (24/4)*(x^5/x^5)*(y^8)------> (6)*(1)*(y^8)-----> 6y^8

the answer is
B: 6y^8
8 0
2 years ago
Recall that the primes fall into three categories: Let Pi be the set of
a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n

(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)

Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1

(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite

4 0
3 years ago
PLEASE HELP ITS MY TRIMESTER TEST MY MOM WILL KILL ME IF I FAIL! &lt;3 WILL MARK BRAINLIST!
Artyom0805 [142]

Answer:

r =2.9985yd

Step-by-step explanation:

Given that the circumference of a circle is 2\pir

where r is the radius and \pi = 3.1415926....

You have

2\pir = 18.84 yd

2(3.1415926)r = 18.84 yd

6.2831r = 18.84yd

divide through by 6.2831

r = 2.9985yd = aprox 3yds

7 0
2 years ago
Find each measure cause i don't understand?
vekshin1

Answer:

x=6 .  m<PQS=82  m<SQR=61 :)

Step-by-step explanation:

(13x+4) + (10x-1) = 141

combine like terms

23x+3=141

subtract 3 from both sides

23x=138

divide both sides by 23

x=6

substitute x into both original equations

m<PQS=13(6)+4

m<PQS=78+4

m<PQS= 82

m<SQR=10(6)+1

m<SQR=60+1

M<SQR=61

4 0
3 years ago
Read 2 more answers
(3^2+4−2)(2^2−3−4) can you guys please solve this for me please.
Degger [83]

Answer:

-33

Step-by-step explanation:

(3^2+4-2)(2^2-3-4)

(9+4-2)(4-3-4)

(13-2)(1-4)

(11) · (-3)

(-33)

6 0
3 years ago
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