Part a)
for extraneous solution
<span>
1 ⋅ sqrt(x+2) + 3 = 0</span>
for non extraneous solution
1⋅ sqrt(x+2) + 3 = 6
part b) solve each equation
1⋅sqrt(x+2)+3=0
x+sqrt(x+2)=−3
square both sides
(sqrt(<span>x+2)</span>)^ 2 = (−3)^2
x+2=9
x=9−2
x=7
do you see why its extraneous
Answer:
<h3>X = 3/5 </h3>
3/5x + 3 = 6
3/5x = 6 - 5
3/5x = 1
<h3><u>Multiplying </u><u>both </u><u>the </u><u>sides </u><u>with </u><u>5x </u></h3>
3/5x (5x) = 1* 5x
3 = 5x
3/5 = X
Answer:
<u>Part A</u>
- Reflect over the y-axis: (x, y) → (-x, y)
A (-4, 4) → (4, 4)
B (-2, 2) → (2, 2)
C (-2, -1) → (2, -1)
D (-4, 1) → (4, 1)
- Shift 4 units down: (x, y-4)
(4, 4-4) → A' (4, 0)
(2, 2-4) → B' (2, -2)
(2, -1-4) → C' (2, -5)
(4, 1-4) → D' (4, -3)
<u>Part B</u>
Two figures are <u>congruent</u> if they have the same shape and size. (They are allowed to be rotated, reflected and translated, but not resized).
Therefore, ABCD and A'B'C'D' are congruent. They are the same shape and size as they have only be reflected and translated.
Answer:
0 (zero)
Step-by-step explanation:
Plug in the variable for -4 and solve
y= 2(-4)2-4(-4)
y= -8(2)-4(-4)
y=-16+16
y=0