Answer:
y=81
Step-by-step explanation:
x=9 and y=81
y-70=11 (add 70 onto 11)
y=81
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1,
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
<span>
</span>
Answer:

- Multiply 5 by 5 to get your first parameter.

- Multiply 6 by 5 to get the denominator, or your second parameter.

- For the second fraction,
, you need to multiply both parameters by 2, similar to before, but we now must use a different number, otherwise, the denominators will not be the same.


- The last step is to put these numbers you gathered into fractions. The bigger number always goes on the bottom, referred to as the denominator, while the smaller number, referred to as the numerator, always goes on the top.


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Finally, the problem is solved. Now that the problem is solved, we review what we just learned <em>not through more problems, though.</em>
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<h3>What have we learned?</h3>
We learned how to efficiently make fractions' deominators match.
Questions related to this topic? Ask me in the comments box, please!
Answer:
125
Step-by-step explanation:
it is the multiflication of 5 by itself 3 times (5×5×5=125)
Answer:

Step-by-step explanation:
take the 3x^2 common