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2 answers:
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Answer:
B) -2x+2y=-2
Step-by-step explanation:
we have
----> equation A
----> equation B
Solve the system by graphing
Remember that the solution of the system of equations is the intersection point both graphs
using a graphing tool
The solution is the point (6,5)
see the attached figure
Remember that
If a equation is added to the system so that the solution does not change, then the solution of the system must be a solution of the equation added
<u><em>Verify each case</em></u>
The solution of the system is (6,5)
substitute the value of x and the value of y in each equation
case A) x-y=2
6-5=2
1=2 ----> is not true
therefore
This equation can't be added to the system
case B) -2x+2y=-2
-2(6)+2(5)=-2
-12+10=-2
-2=-2 ----> is true
therefore
This equation can be added to the system
case C) 3x+y=20
3(6)+5=20
18+5=20
23=20 ----> is not true
therefore
This equation can't be added to the system
case D) x+2y=18
5+2(6)=18
5+12=18
17=18 ----> is not true
therefore
This equation can't be added to the system
We need to determine whether is a terminating decimal or a repeating decimal.
Let's solve this question using the long division method
First, let's identify the divisor and dividend. The number to be divided is 11 hence this is the dividend, and it needs to be divided by 128 which is the divisor
Next, since the divisor (128) is greater than the dividend (11) it can not divide 11. Hence, we will introduce a decimal point in quotient, and append a 0 next to 11 and divide 110 by 128. Again, 128 is greater than 110 so we will introduce a 0 in the quotient, and append another 0 next to 110, and will divide 1100 by 128. We will see what multiple of 128 is less than or equal to 1100. That multiple is 8. So we write 8 in the quotient and multiply 128 with 8 and subtract the product (128*8 = 1024) from 1100. The remainder that we get is 76.
Next, we append a 0 to the remainder and divide 760 by 128. Now, we see what multiple of 128 is less than or equal to 760. That multiple is 5. So we write 5 next to the quotient and multiply 128 with 5 and subtract the product (640) from 760. Now, the remainder is 120.
Next, we append a 0 to the remainder and divide 1200 by 128. Now, we see what multiple of 128 is less than or equal to 1200. That multiple is 9. So we write 9 next to the quotient and multiply 128 with 9 and subtract the product (1152) from 1200. Now, the remainder is 48.
Next, we append a 0 to the remainder and divide 480 by 128. Now, we see what multiple of 128 is less than or equal to 480. That multiple is 3. So we write 3 next to the quotient and multiply 128 with 3 and subtract the product (384) from 480. Now, the remainder is 96.
Next, we append a 0 to the remainder and divide 960 by 128. Now, we see what multiple of 128 is less than or equal to 960. That multiple is 7. So we write 7 next to the quotient and multiply 128 with 7 and subtract the product (896) from 960. Now, the remainder is 64.
Next, we append a 0 to the remainder and divide 640 by 128. Now, we see what multiple of 128 is less than or equal to 640. That multiple is 5. So we write 5 next to the quotient and multiply 128 with 5 and subtract the product (640) from 640. Now, the remainder is 0.
Hence, we have solved the entire problem
Last, we look at the quotient i.e. 0.0859375, which is the solution to the problem. We see that the quotient has a definite number of digits in it, and terminates at 5. Hence, this is a terminating decimal.
A repeating decimal is one in which a particular pattern after the decimal point keeps re-occuring, which is not the case here. Hence, is a terminating decimal.
Please refer to the attached image for visualization
