|DF| = |DE| + |EF|
|DF| = 9x -36
|DE| = 47
|EF| = 3x+10
Substitute:
9x - 39 = 47 + 3x + 10
9x - 39 = 3x + 57 |+39
9x = 3x + 96 |-3x
6x = 96 |:6
x = 16
Put the value of x to the equation |EF| = 3x + 10
|EF| = (3)(16) + 10 = 48 + 10 = 58
Answer: |EF| = 58
Answer:
Step-by-step explanation:
You need to complete the square.
C(x) = 0.02(x^2 - 1000x ...) + 11000
C(x) = 0.02 (x^2 - 1000x + 500^2) + 11000 - 5000
C(x) = 0.02 (x^2 - 1000x + 500^2) + 6000
C(x) = 0.02(x - 500)^2 + 6000
Now if you look at the answer you will find that the square is completed. That means that number of tractors you could produce is 500 at a cost of 6000
There is a flow to this question that you may have trouble understanding.
First of all the 500^2. That comes from taking 1/2 of 1000 and squaring it. That's what you need to complete the square.
Bur that is not what you have adding into the equation. Remember that there is a 0.02 in front of the brackets.
500^2 = 250000
0.02 * 250000 = 5000
So that number must be subtracted to make the square = 0. When you remove the brackets, you should get 11000 all in all.
So what you have outside the brackets is 11000 - 5000 = 6000
The rest is just standard for completing the square.
Given:
The table of values is
Number of Students : 7 14 21 28
Number of Textbooks : 35 70 105 140
To find:
The rate of change and showing that the ratios of the two quantities are proportional and equivalent to the unit rate.
Solution:
The ratio of number of textbooks to number of students are
All the ratios of the two quantities are proportional and equivalent to the unit rate.
Let y be the number of textbooks and x be the number of students, then
Here, k=5.
Hence the rate of change is constant that is 5.
Answer:
Step-by-step explanation:
The length of the midsegment is half of the base in this case.
12x = 8x + 40
4x=40
x=10
So it's 60
Answer:
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Step-by-step explanation:
