<span>1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-2);y=-x-2
D.y=-x
2.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-1);y=-3/2x-6
C.y=-3/2x+2
3.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (4,2);x=-3
D.y=4
4.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (-2,3);y=1/2x-1
B.y=-2x-1
5.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (5,0);y+1=2(x-3)
D.y=-1/2x+5/2</span>
Answer:
the answer is c=34
Step-by-step explanation:
use the pythogoras therom since there is a right angle triangle c2=a2+b2
hope this helped:)
3x-5=19-x
3x-5-19+x=0
4x-24=0
4x=24
x=6
Answer:
l=0.1401P\\
w =0.2801P
where P = perimeter
Step-by-step explanation:
Given that a window is in the form of a rectangle surmounted by a semicircle.
Perimeter of window =2l+\pid/2+w
Or 
To allow maximum light we must have maximum area
Area = area of rectangle + area of semi circle where rectangle width = diameter of semi circle
Hence we get maximum area when i derivative is 0
i.e. 
Dimensions can be
Diameter of the circle is BD as shown, as indicated and passed through center where BD passes, triangle ABD is a right triangle with angle ABD as the right angle. It means ADB and ABD are complementary which means
AB = 140 degree
AD=40 degree
