<h2>
Step-by-step explanation:</h2>
As per the question,
Let a be any positive integer and b = 4.
According to Euclid division lemma , a = 4q + r
where 0 ≤ r < b.
Thus,
r = 0, 1, 2, 3
Since, a is an odd integer, and
The only valid value of r = 1 and 3
So a = 4q + 1 or 4q + 3
<u>Case 1 :-</u> When a = 4q + 1
On squaring both sides, we get
a² = (4q + 1)²
= 16q² + 8q + 1
= 8(2q² + q) + 1
= 8m + 1 , where m = 2q² + q
<u>Case 2 :-</u> when a = 4q + 3
On squaring both sides, we get
a² = (4q + 3)²
= 16q² + 24q + 9
= 8 (2q² + 3q + 1) + 1
= 8m +1, where m = 2q² + 3q +1
Now,
<u>We can see that at every odd values of r, square of a is in the form of 8m +1.</u>
Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.
Hence , it is clear that square of an odd positive is in form of 8m +1
His brothers have 4 games
He had 10 games
So in total there are 14 games
And if there are 2 cases and each one had the same amount, then in each one there would be 14÷2=7
In each case, there are 7 games.
Answer:Yes
Step-by-step explanation:
The value of x is ![5\sqrt[3]{25}](https://tex.z-dn.net/?f=5%5Csqrt%5B3%5D%7B25%7D)
.
Solution:
Given expression is ![-7=8-3 \sqrt[5]{x^{3}}](https://tex.z-dn.net/?f=-7%3D8-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D)
.
Switch both sides.
![8-3 \sqrt[5]{x^{3}}=-7](https://tex.z-dn.net/?f=8-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%3D-7)
Subtract 8 from both side of the equation.
![8-3 \sqrt[5]{x^{3}}-8=-7-8](https://tex.z-dn.net/?f=8-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D-8%3D-7-8)
![-3 \sqrt[5]{x^{3}}=-15](https://tex.z-dn.net/?f=-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%3D-15)
Divide by –3 on both side of the equation.
![$\frac{-3 \sqrt[5]{x^{3}}}{-3} =\frac{-15}{-3}](https://tex.z-dn.net/?f=%24%5Cfrac%7B-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%7D%7B-3%7D%20%3D%5Cfrac%7B-15%7D%7B-3%7D)
![\sqrt[5]{x^{3}}=-5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%3D-5)
To cancel the cube root, raise the power 5 on both sides.
![(\sqrt[5]{x^{3}})^5=(-5)^5](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%29%5E5%3D%28-5%29%5E5)
To find the value of x, take square root on both sides.
![\sqrt[3]{x^3}=\sqrt[3]{25}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%3D%5Csqrt%5B3%5D%7B25%7D)
![x=5\sqrt[3]{25}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B25%7D)
Hence the value of x is .
Answer:
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