They are the same slope
they are negative inversees (they multily to get -1)
2
-1/2
use the square viewer (on TI)
The relationship between the slopes of two lines that are parallel is they are the same.
The relationship between the slopes of two lines that are perpendicular is they are negative inverses of each other (they multiply to -1).
A line that is parallel to a line whose slope is 2 has slope 2.
A line that is perpendicular to a line whose slope is 2 has slope -1/2.
What must be done to make the graphs of two perpendicular lines appear
to intersect at right angles when they are graphed using a graphing
utility?
Answer:
None of the above
Step-by-step explanation:
3^3=81
-3^3=-81
6^3=216
-6^3=-216
5^3=125
Firstly, the rate of increase of Canton = 80/7720 = 0.010
The rate of increase of HP = 120/3200 = 0.0375
Apply the formula of compound interest==> A =A'(1+i)^n
The population will be equal when the following equality is satisfied
7720(1+0.01)^n = 3200(1+0.0375)^n
or
7720x1.01^n = 3200x1.0375^n
Divide both sides by 3200 ===>(193/80)x1.01^n=1.0375^n
Answer:
127.17cm²
Step-by-step explanation:
Area of a semicircle: πr² ÷ 2
d = 18
r = 18/2 = 9
Area of a semicircle: 9²π ÷ 2 = 127.17cm²
Answer:
<em>It has been given that Rectangle Q has an area of 2 square units.</em>
<em>Thea Drew a scaled version of Rectangle Q and marked it as R.</em>
<em>As you must keep in mind If we draw scaled copy of pre-image, then the two images i.e Pre-image and Image are similar.</em>
<em>As you have not written what is the scale factor of transformation</em>
<em>Suppose , Let the Scale factor of transformation= k</em>
<em>Rectangle Q = Pre -image, Rectangle R= Image</em>
<em>If, Pre-Image < Image , then scale factor is k >1.</em>
<em>But If, Pre-Image > Image , then Scale factor will be i.e lies between, 0<k<1.</em>
