AB=48, DC=88
48+88=136
136÷2=68
Answer: LM=68
Remember that the length of the mid segment in a trapezoid is half the sum of the base lengths.
Option C
For each value of y, -2 is a solution of -21 = 6y - 9
<u>Solution:</u>
Given, equation is – 21 = 6y – 9
We have to find that whether given set of options can satisfy the above equation or not
Now, let us check one by one option
<em><u>Option A) </u></em>
Given option is -5
Let us substitute -5 in given equation
- 21 = 6(-5) – 9
- 21 = -30 – 9
- 21 = - 39
L.H.S ≠ R.H.S ⇒ not a solution
<em><u>Option B)</u></em>
Given option is 3
- 21 = 6(3) – 9
- 21 = 18 – 9
- 21 = 9
L.H.S ≠ R.H.S ⇒ not a solution
<em><u>Option C)</u></em>
Given option is -2
- 21 = 6(-2) – 9
- 21 = - 12 – 9
- 21 = - 21
L.H.S = R.H.S ⇒ yes a solution
<em><u>Option D)</u></em>
- 21 = 6(9) – 9
- 21 = 54 – 9
- 21 = 45
L.H.S ≠ R.H.S ⇒ not a solution
Hence, the solution for the given equation is – 2, so option c is correct
There are 600 students including the seventh and eighth graders at the party.
This problem uses the concept of percentages to define the conditions that are laid in front of us.
Let the original number of students be S , and the number of seventh graders be = 0.60S
We know that percent is used to convey the mathematical term of a fraction multiplied by 100.
Total students after 20 eighth graders arrive = S + 20
And we have that
Number of seventh graders / total number of students = 58%
.60S / [ S + 20 ] = .58 we multiply both sides by S + 20
0.60S =0 .58 [ S + 20]
.60S = .58S + 11.6 we subtract 0.58S from both the sides
0.02S = 11.6 we divide both the sides by .02
S = 11/6 / 0.02 = 580
So the total number of students = 580 + 20 = 600 .
Hence there are 600 students at the party at that time.
To learn more about students visit:
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