Hi there, (6z²+z-1)(9z-5)=(6z²+z+-1)(9z+-5)=(6z²)(9z)+(6z²)(-5)+(z)(9z)+(z)(-5)+(-1)(9z)+(-1)(-5)=54z³-30z²+9z²-5z-9z+5=54z³-21z²-14z+5. Therefore, the answer is 54z³-21z²-14z+5.
Answer:
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Part A.
You need two equations with the same slope and different y-intercepts.
Their graph is parallel lines. Since the lines do not intersect, there is no solution.
y = 2x + 2
y = 2x - 2
Part B.
We use the first equation as above. For the second equation, we use an equation with different slope. Two lines with different slopes always intersect.
y = 2x + 2
y = -2x - 2
In the second equation, y = -2x - 2. We now substitute -2x - 2 for y in the first equation.
-2x - 2 = 2x + 2
-4x = 4
x = -1
Now substitute -1 for x in the first equation to find y.
y = 2x + 2
y = 2(-1) + 2
y = -2 + 2
y = 0
Solution: x = -1 and y = 0
Answer:
Q1. x= 18, y=59
Q2. m∠J= 56°
Step-by-step explanation:
Q1. (3x +5)°= y° (base ∠s of isos. △)
y= 3x +5 -----(1)
(3x +5)° +y° +(4x -10)°= 180° (∠ sum of △)
3x +5 +y +4x -10= 180
7x +y -5= 180
7x +y= 180 +5
7x +y= 185 -----(2)
Substitute (1) into (2):
7x +3x +5= 185
10x= 185 -5
10x= 180
x= 180 ÷10
x= 18
Substitute x= 18 into (1):
y= 3(18) +5
y= 59
Q2. (5x -13)°= (3x +17)° (base ∠s of isos. △)
5x -13= 3x +17
5x -3x= 17 +13
2x= 30
x= 30 ÷2
x= 15
∠LKJ
= 3(15) +17
= 62°
∠KLJ= 62° (base ∠s of isos. △)
m∠J
= 180° -62° -62° (∠ sum of △JKL)
= 56°
