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Dovator [93]
3 years ago
15

Is 2 (5x-1) = 3 (x+11) a one solution or a no solution or a real number

Mathematics
1 answer:
spin [16.1K]3 years ago
5 0

Answer: A one solution.

x=5

Step-by-step explanation: (2)(5x)+(2)(−1)=(3)(x)+(3)(11)                                   10x+−2=3x+33                                                                                                       10x−2-3x=3x+33-3x                                                                                             7x−2+2=33+2                                                                                                       7x/7=35/7                                                                                                               x=5

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JulsSmile [24]

Answer:

206  2/3 ft³

Step-by-step explanation:

The volume of this solid is given by V = (1/3)(base area)(height).

Here, V = (1/3) (62 ft²) (10 ft ) = 206  2/3 ft³

5 0
3 years ago
The sign of the product of 2 integers with the same sign is positive. what is the sign of the product of 3 integers with the sam
klemol [59]
Product of 3 negatives equals a negative answer
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9/10 + 9/100 + 9/1000 + 9/10000 + … =
Korvikt [17]

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3 years ago
Ms. Fitzgerald had 2 1 /4 gallons of fruit punch. She served 3 /8 gallon of the fruit punch
Luda [366]

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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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