Answer:
-8b^3ax^2
Step-by-step explanation:
Answer is b the answer is because there is no C or D in my conversations are in my store
Answer:
im in 8th grade and im getting your questions
Step-by-step explanation:
Answer:
option 4.
16 square units
Step-by-step explanation:
as we do not have the measures of the sides, but if the points of the vertices with Pythagoras we can calculate the sides.
P = (2 , 4)
S = (4 , 2)
we have to subtract the values of p from s
PS = (4 - 2 , 2 - 4)
PS = (2 , -2)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
PS^2 = 2^2 + -2^2
PS = √ 4 + 4
PS = √8
PS = 2√2
S = (4 , 2)
R = (8 , 6)
SR = (8-4 , 6-2)
SR = (4 , 4)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
SR^2 = 4^2 + 4^2
SR = √ (16 + 16)
SR = √32
SR = 4√2
having the values of 2 of its sides we multiply them and obtain their area
PS * RS = Area
2√2 * 4√2 =
16
(-1.2,-2.0) and (1.9,2.2) are the best approximations of the solutions to this system.
Option B
<u>Step-by-step explanation:</u>
Here, we have a graph of two functions from which we need to find the approximate value of common solutions. Let's find this:
First look at where we have intersection points, In first quadrant & in third quadrant.
<u>At first quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point . After doing this we can clearly see that the perpendicular lines cut x-axis at x=1.9 and y-axis at y=2.2. So, one point is (1.9,2.2)
<u>At Third quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point. After doing this we can clearly see that the perpendicular lines cut x-axis at x=-1.2 and y-axis at y= -2.0. So, other point is (-1.2,-2.0).
