Because this is a second degree equation you will have 2 solutions. When you take the square root of a number you have to account for both the positive and negative roots. Since the square root of 289 is 17 then your solutions are +17 and -17.
Answer:
x = 0
y = 5
z = -5.
Step-by-step explanation:
X-5y-z=-20
-2x-y-6z=25
-x-4y-z=-15
Add equation 1 and 3:
-9y - 2z = -35 .......... A
Multiply the first equation by 2:
2x - 10y - 2z = -40
Now add this to equation 2:
-11y -8z = -15.............B
Now multiply equation A by -4:
36y + 8x = 140.........C
Add equations B and C:
25y = 125
y = 5.
Substitute for y in equation A:
-9*5 - 2z = -35
2z = -45 + 35 = - 10
z = -5.
Now find x by substituting for y and z in the first equation:
x - 5(5) - (-5) = -20
x - 20 = -20
x = 0.
If there was 1 green for every 3 there would be 6 picks in total
Step-by-step explanation:
-3x² + 7x + 2 + x³ - x² -9
= x³ -3x² -x² + 7x +2 -9
= x³ -4x² + 7x -7
The first step for solving this expression is to remove the first set of parenthesis.
8j³ - 10j² - 7 - (6j³ - 10j² - j + 12)
When there is a "-" sign in front of the parenthesis,, you must change the sign of each term in the parenthesis. This will look like the following:
8j³ - 10j² - 7 - 6j³ + 10j² + j - 12
Eliminate the opposites in the expression
8j³ - 7 - 6j³ + j - 12
Collect the first set of like term (ones with j³)
2j³ - 7 + j - 12
Calculate the difference between basic numbers
2j³ - 19 + j
Lastly,, use the commutative property to reorder the terms
2j³ + j - 19
Since you cannot simplify this expression any further,, the correct answer to your question is 2j³ + j - 19.
Let me know if you have any further questions
:)
