It will be 0.073. because the 73 is in the 1000 place
Solid...(-3,0)(-4,-1)...shaded below the line
slope = (-1-0) / (-4-(-3) = -1/(-4 + 3) = -1/-1 = 1
y = mx + b
-1 = 1(-4) + b
-1 = -4 + b
-1 + 4 = b
3 = b
this line is : y < = x + 3 <== (thats less then or equal)
====================
solid...(1,1)(2,-1)...shaded below the line
slope = (-1-1) / (2-1) = -2/1 = -2
y = mx + b
1 = -2(1) + b
1 = -2 + b
1 + 2 = b
3 = b
this line is : y < = -2x + 3 <== (thats less then or equal)
A Kilometer is 1,000 times larger than a meter
Add
5
5
to both sides of the equation.
√
2
x
+
13
=
x
+
5
2
x
+
13
=
x
+
5
To remove the radical on the left side of the equation, square both sides of the equation.
(
√
2
x
+
13
)
2
=
(
x
+
5
)
2
(
2
x
+
13
)
2
=
(
x
+
5
)
2
Simplify each side of the equation.
2
x
+
13
=
x
2
+
10
x
+
25
2
x
+
13
=
x
2
+
10
x
+
25
Solve for
x
x
.
x
=
−
2
,
−
6
x
=
-
2
,
-
6
Exclude the solutions that do not make
√
2
x
+
13
−
5
=
x
2
x
+
13
-
5
=
x
true.
x
=
−
2
The right answer is:
<em>area B = area C</em>
We can solve this problem by using Kepler's laws of planetary motion. There are three Kepler's laws. In this exercise, we need to use the second law. According to this law,<em> a line segment joining a planet and the sun sweeps out equals areas during equals intervals of time. </em>So, a certain planet sweeps out an <em>area B </em>from the point <em>P3 </em>to <em>P4</em> in an interval of time <em>t. </em>On the other hand, for the same interval of time <em>t, </em>the planet sweeps out an <em>area C </em>from point <em>P4</em> to <em>P5, </em>that is equal to the previous area according to second kepler's law.
