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NARA [144]
3 years ago
11

Which of these numbers are less than 8.1 × 10^-8?

Mathematics
1 answer:
lina2011 [118]3 years ago
8 0
The answer is 3.4×10^-10
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A survey of 350 people found 150 like birds. 170 like squirrels and 30 like both animals. How many like neither?
Arada [10]

Answer:

0

Step-by-step explanation:

150 + 170 + 30 = 350

170 + 30 = 200

200 + 150 = 350

Hope this helps! :)

8 0
3 years ago
Which of the following shapes are rectangles ? Please choose 2 correct answers !!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!
timurjin [86]

Answer:

I think it is b and d .

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is divide 6/13 by 6/12
Brut [27]
I believe it is 12/13
5 0
3 years ago
Andre is paid $12 per hour at his caddying job and $9.50 per hour working at a local fast food restaurant. He would like to keep
Fynjy0 [20]

The equation that can be used to model this scenario is 12x + 9.5y ≥ 275 and x + y ≤ 25

Let x represents the number of hours Andre works at his caddying job and y represents the number of hours he works at the restaurant.

Since he earn at least $275.00 a week, hence:

12x + 9.5y ≥ 275   (1)

Also, he can work no more than 25 hours, hence:

x + y ≤ 25   (2)

Therefore the equation that can be used to model this scenario is 12x + 9.5y ≥ 275 and x + y ≤ 25

Find out more at: brainly.com/question/25285332

5 0
3 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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