The solutions to the given system of equations is (0, -6) and (1, -5)
<h3>Simultaneous equations</h3>
From the question, we are to determine the solutions to the given system of equations
The equations are
x − y = 6 --------- (1)
y = x² −6 ---------- (2)
From equation (1)
x - y = 6
∴ x = 6 + y ------- (3)
Substitute into equation (2)
y = x² −6
y = (6+y)² −6
y = (6+y)(6+y) -6
y = 36 + 6y + 6y +y² -6
y = 36 + 12y + y² - 6
Simplifying
y² + 12y - y + 30 = 0
y² + 11y + 30 = 0
Solve quadratically
y² + 11y + 30 = 0
y² + 6y + 5y + 30 = 0
y(y +6) +5 (y +6) = 0
(y + 5)(y + 6) = 0
y + 5 = 0 OR y + 6 = 0
y = -5 OR y = -6
Substitute the values of y into equation (3)
x = 6 + y
When y = -5
x = 6 + (-5)
x = 6 -5
x = 1
When y = -6
x = 6 + (-6)
x = 6 -6
x = 0
∴ When x = 0, y = -6 and when x = 1, y = -5
Hence, the solutions to the given system of equations is (0, -6) and (1, -5)
Learn more on Solving simultaneous equations here: brainly.com/question/16863577
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Answer:
84 is the highest possible course average
Step-by-step explanation:
Total number of examinations = 5
Average = sum of scores in each examination/total number of examinations
Let the score for the last examination be x.
Average = (66+78+94+83+x)/5 = y
5y = 321+x
x = 5y -321
If y = 6, x = 5×6 -321 =-291.the student cannot score -291
If y = 80, x = 5×80 -321 =79.he can still score higher
If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.
If y= 100
The average cannot be 100 as student cannot score 179(maximum score is 100)
Because the situation represents a profit, the integer is 9.
Write it as a fraction
11x + 32
X + 3
Answer:
<h2>The answer is - 16</h2>
Step-by-step explanation:
<h3>
</h3>
where
n is the number of terms
From the question since we are finding the 5th term
n = 5
Substitute the value into the above formula and solve
We have
We have the final answer as
<h3>- 16</h3>
Hope this helps you
