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aliina [53]
3 years ago
14

The sum of two numbers is 8, and their product is 80.

Mathematics
1 answer:
Anettt [7]3 years ago
6 0

From their sum, the numbers are ' x ' and ' 8 - x ' .

From their product,  x(8 - x) = 80 .

When you massage and manipulate this equation into the standard
quadratic form, you have

         x² - 8x + 80 = 0  .

Applying the quadratic formula to tease out the solutions of this equation,
we find that they are

          <em> x = 4 + 64 i</em>
and
          <em> x = 4 - 64 i</em>


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jek_recluse [69]
What the heckk??????????
6 0
3 years ago
Please help me :):)):):
Verdich [7]

Answer:

12

Step-by-step explanation:

5^2 + b^2 = 13^2

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answer: 12

3 0
3 years ago
If Joey had 5,000,000 dollars and spent 28% of the money, how much will he have left?
Mrac [35]

Answer:

3600000

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3 years ago
Are the two triangles below similar??
kodGreya [7K]

Answer:

Yes

Step-by-step explanation:

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7 0
2 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

5 0
3 years ago
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