A(4,2) to (X + 1, y - 5)
so
A' (5, -3)
then reflected across y-axis
A' (-5, -3)
a)A(4,2), point A is located in quadrant I
b)A'(5,-3), point A' is located in quadrant IV
c)A'(-5,-3), point A' is located in quadrant III
2x+1>x-(1-x)
2x+1>x-1+x
2x+1>2x-1
1>-1
Any value of x will make this true.
Answer:
0
Step-by-step explanation:
Because anything times 0 is 0
Option 1 : -22
using formula
(a + b )(a - b ) = a^2 - b^2
(√10 + 2√8) (√10 - 2√8)
= (√10)^2 - (2√8)^2
= 10 - 2×2×8=10-32
= -22
2nd question answer is option 3
Answer:
(e) 1 =64⁰
(f) 1/64 = 64⁻¹ negative one power
(g) 1/8 = 64⁻¹/² negative one half power
(h) 1/2 = 64⁻1/6 negative one/sixth power
(I) 1/64 same as (f) ??
Step-by-step explanation:
