So 1/9-16z^2 this is a diffirence of two perfect squares thing
so (1/3)^2-(4z)^2
so ((1/3)-4z)((1/3)+4z)
The sum of length and width is half the perimeter, 160 yd. The length is the average of the sum and difference of dimensions:
.. (160 +20)/2 = 90
Then the width is 160 -90 = 70
The rectangular field is 70 yd by 90 yd.
Answer: the distance of the base of the house to the foot of the ladder is 6.84 feet
Step-by-step explanation:
The scenario is shown in the attached photo.
Right angle triangle ABC is formed when the ladder leans against the wall of the house.
AC = the height of the ladder
AB = x feet = distance of the base of the house to the foot of the ladder
BC is the wall of the building.
To determine x, we will apply trigonometric ratio
Cos # = adjacent/hypotenuse
Where
# = 70 degrees
Hypotenuse = 20
Adjacent = x
Cos 70 = x/20
x = 20cos70
x = 20 × 0.3420
x = 6.84 feets
Answer:
no solution
Step-by-step explanation:
y = -2x + 3
6x + 3y = -3
the substitution method means you plug one equation into the next, because the first equation gives us a solution for y we can go ahead and plug that into y of the second equation
6x + 3(-2x + 3) = -3
6x - 6x + 9 = -3
9 = -3
which is false meaning that there are no solutions and the lines don't touch at any point