<span>System 1 and system 2, because the second equation in system 2 is obtained by adding the first equation in system 1 to two times the second equation in system 1
This is the correct answer because not only is it true but it also follows the property of solving systems of equations with adding the equations. To prove that it is true:
2nd equation in system #2 = 1st equation in system #1 + 2(2nd equation in system #1)
</span>10x − 7y = 18 == 4x − 5y = 2 + 2(<span>3x − y = 8)
10x - 7y = 18 == 4x - 5y = 2 + 6x - 2y = 16
10x = 7y = 18 == 10x - 7y = 18</span>
Alright, since there are 5 numbers, and the mean (or average) is (sum)/(amount of numbers), we have (sum)/5=14. Multiplying both sides by 5, we have the sum being 80. The median of 10 means that in a, b, c, d, e, 10 has to be c and the numbers have to be in ascending order. A and b must be 10 or lower, while d and e must be 10 or higher. Putting some random numbers in, we can have 1, 1, 10, 15, and e. We left e there because the sum needs to be 80, and since 1+1+10+15=27, 80-27=53=e. This, however, would not work if e was less than 10 and we therefore would have needed to make some numbers lower to compensate for this. Our answer is therefore 1, 1, 10, 15, 53
Answer:
7
Step-by-step explanation:
The greatest common factor or GCF of 7 and 49 is 7 because 7 is the highest number that can go into both numbers 7 and 49. No other number can go into those numbers except 7.
The solutions of the function are:
- When f(x) = -4, the solution is -4
- When f(x) = -2, the solution is -5/2
- When f(x) = 0, the solution is -1
<h3>How to solve for the equation </h3>
The equation is given as
When x = -4
3/4 * -4 -1
= -12/4 - 1
= -4
When x = -2
3/4*(-4) - 1
= -6/4 - 1/1
Take the lcm
-10/4
= -5/2
When x = 0
3/4(0) - 1
= -1
When f(x) = -4, the solution is -4
When f(x) = -2, the solution is -5/2
When f(x) = 0, the solution is -1
Read more on real numbers here:
brainly.com/question/155227
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Answer:
0.10 or 10%
Step-by-step explanation:
Since each student must take at least 2 courses, if no student took all three courses, the sum of the percentages would total 200% (counting each student twice). Therefore, the proportion of students taking all three courses is the difference between the sum of all percentages and 200%:
0.10 or 10% of the students take all 3 courses.
