answer.
Answer:
x=2 and y=0 is the required result.
Step-by-step explanation:
We have been given system of equations:
5x+2y=105x+2y=10 (1)
And 3x+2y=63x+2y=6 (2)
We will use elimination method:
Multiply 1st equation by 3 and 2nd equation by 5 we get:
15x+6y=3015x+6y=30 (3)
15x+10y=3015x+10y=30 (4)
Now subtract (4) from (3) we get:
-4y=0−4y=0
y=0y=0
Now, put y=0 in (1) equation:
5x+2(0)=105x+2(0)=10
5x=105x=10
x=2x=2
Hence, x=2 and y=0
Answer:
-5
Step-by-step explanation:
-4x +10 =5(x +11)
-4x +10 =5x +55
-4x - 5x =55 - 10
-9x =45
x=-45 :9
x=-5
-6 - 3 = -9
12 - (-6) = 18
4 - 12 = -8
20 - 4 = 16
The first difference is -9. The third difference is -8.
Maybe the fifth difference is -7.
That makes the next term 13.
<span>3, -6, 12, 4, 20, 13</span>
Answer:
q = 15
Step-by-step explanation:
Given
f(x) = x² + px + q , then
f(3) = 3² + 3p + q = 6 , that is
9 + 3p + q = 6 ( subtract 9 from both sides )
3p + q = - 3 → (1)
---------------------------------------
f'(x) = 2x + p , then
f'(3) = 2(3) + p = 0, that is
6 + p = 0 ( subtract 6 from both sides )
p = - 6
Substitute p = - 6 into (1)
3(- 6) + q = - 3
- 18 + q = - 3 ( add 18 to both sides )
q = 15
