This is the diagram
Answer:
b=9
Step-by-step explanation:
This photo is the diagram now for the solution
using the Pythagoras theorem
we know that
C²=a²+b²
now we are looking for the third side we are given the value of A which is 6 and C which is 3√13 then that suppose that we are looking for b
we then make b the subject of the formula
C²=A²+B²
B²=C²-A²
B=√C²-√A²
then we substitute for the values
B=√(3√13)²-√6²
B=√117-√36
B=√81
B=9
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X = is How many ex’s and O’s do you have?
2 is,I see
3 is,how you do
4 is, LOVE
Y= ask first
11, is make a wish
16.5, divide by 4. Then you define what remains 5
2+2=4 Love again
Answer:
the two roots are x = 1 and x = 4
Step-by-step explanation:
Data provided in the question:
(x³ − 64) (x⁵ − 1) = 0.
Now,
for the above relation to be true the following condition must be followed:
Either (x³ − 64) = 0 ............(1)
or
(x⁵ − 1) = 0 ..........(2)
Therefore,
considering the first equation, we have
(x³ − 64) = 0
adding 64 both sides, we get
x³ − 64 + 64 = 0 + 64
or
x³ = 64
taking the cube root both the sides, we have
∛x³ = ∛64
or
x = ∛(4 × 4 × 4)
or
x = 4
similarly considering the equation (2) , we have
(x⁵ − 1) = 0
adding the number 1 both the sides, we get
x⁵ − 1 + 1 = 0 + 1
or
x⁵ = 1
taking the fifth root both the sides, we get
![\sqrt[5]{x^5}=\sqrt[5]{1}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E5%7D%3D%5Csqrt%5B5%5D%7B1%7D)
also,
1 can be written as 1⁵
therefore,
![\sqrt[5]{x^5}=\sqrt[5]{1^5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E5%7D%3D%5Csqrt%5B5%5D%7B1%5E5%7D)
or
x = 1
Hence,
the two roots are x = 1 and x = 4
Answer:
The answer to your question is: 5⁵
Step-by-step explanation
5³ x 25
First, find the prime factors of 25
25 5
5 5
1 so, 25 = 5 x 5 or 5²
5³ x 5² Remeber that final power is the sum of the powers
5⁵ Finally
