Answer:
- ) x = ±4
- ) t = ±9
- ) r = 10
- ) x = ±12
- ) s = ±5
- ) x = ±[tex]\sqrt{255}/2[/tex]
- ) h = ±7
- ) x = 17, -9
- ) s = 128
- ) k = ±3∫
Solutions:
For no. 1
- Take the square root of both sides.
x = ±[tex]\sqrt{16}[/tex]
- Since 4 × 4 = 16, the square root of 16 is 4.
x = ±4
For no. 2
- Take the square root of both sides.
t = ±[tex]\sqrt{81}[/tex]
- Since 9 × 9 = 81, the square root of 81 is 9.
t = ±9
For no. 3
r² = 100
- Take the square root of both sides.
r = ±[tex]\sqrt{100}[/tex]
- Since 10 × 10 = 100, the square root of 100 is 10.
r = ±10
For no. 4
x² = 144
- Take the square root of both sides.
x = ±[tex]\sqrt{144}[/tex]
- Since 12 × 12 = 144, the square root of 144 is 12.
x = ±12
For no. 5
s² = 50/2
s² = 25
- Take the square root of both sides.
s = ±[tex]\sqrt{25}[/tex]
- Since 5 × 5 = 25, the square root of 25 is 5.
s = ±5
For no. 6
4x² = 255
x² = 255/4
- Take the square root of both sides.
x = ±[tex]\sqrt{255/4}[/tex]
- Simplify [tex]\sqrt{255/4}[/tex] to [tex]\sqrt{255/4}[/tex].
x = ±[tex]\sqrt{255\sqrt{4} }[/tex]
- Since 2 × 2 = 4, the square root of 4 is 2.
x = ±[tex]\sqrt{255/2}[/tex]
For no. 7
3h² = 147
h² = 147/3
h² = 49
- Take the square root of both sides.
h = ±[tex]\sqrt{49}[/tex]
- Since 7 × 7 = 49, the square root of 49 is 7.
h = ±7
Other solutions are in pictures in order.
- The first picture is for no. 8
- The second picture is for no. 9
- The last picture is for no. 10
