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Sergeeva-Olga [200]
3 years ago
12

Please solve this problem... I will mark you brainliest

Mathematics
1 answer:
Blababa [14]3 years ago
8 0
I hope this helps you

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HELP ME PLEASE REALLY IMPORTANT OR ILL GET IN TROUBLE!! WILL MARK AS BRAINLEST IF YOU GET IT CORRECT!!
sammy [17]

Answer:

7

Step-by-step explanation:

3 0
3 years ago
Eric bought a ruler for $1.29 and an eraser for $0.39. He paid with $2.00. What is the correct amount of change he should get ba
igor_vitrenko [27]

Answer:

$0.32, so a quarter a nickel and 2 pennies

Step-by-step explanation:

8 0
3 years ago
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A single die is rolled. How many ways can you roll a number that is prime, followed by a 6?
Grace [21]

Answer:

3 ways

Step-by-step explanation:

Since 6 will remain constant throughout the testing, we just need to find all prime numbers 1-6.

1 - is not prime nor composite

2 - is prime

3 - is prime

4 - 2x2=4, so composite

5 - is prime

6 - 2x3=6, so composite

Therefore, 2, 3, and 5 are prime numbers, and there are 3 of them.

5 0
3 years ago
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As a student you are able to earn extra money by assisting your neighbors with odd jobs. If you charged $10.25 an hour for your
dsp73
Okay. so, to find the answer for this, you should set the equation as 10.25x = 8,425. "x" equals the number of hours. What you should do is divide each side by 10.25 to isolate the "x". When you do that 8,425/10.25, you get 821.951 and other numbers or 822 hours when rounded to the nearest whole number. x is approximately 822. You would have to work about 822 hours to earn $8,425.
6 0
3 years ago
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Let E = {(x, y) e R2|xy > 0}. Determine whether E is a subspace of R2 . X3
gayaneshka [121]

Answer:

E is not a subspace of \mathbb{R}^2

Step-by-step explanation:

E is not a subspace of  \mathbb{R}^2

In order to see this, we must find two points (a,b), (c,d) in  E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.  

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

5 0
3 years ago
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