Step 1) Draw a line from point F to point S. A rectangle forms (rectangle FSCW)
Step 2) Find the area of this rectangle. The area is 18 square units because it is 2 units high and 9 units across (9*2 = 18). You can count out the spaces or you can note how we go from x = -4 to x = 5 so subtracting the values gives -4-5 = -9 which has an absolute value of 9.
Step 3) Find the area of triangle FSN. The base is 9 units and the height is 6 units (count out the spaces or subtract y values). So the area is A = b*h/2 = 9*6/2 = 54/2 = 27
Step 4) Add up the area of the rectangle to the area of the triangle: 18+27 = 45
Final Answer: 45 square units
note: another way to find the answer is to find the area of rectangle WABC where point A is at (-4,4) and point B is at (5,4). Then subtract off the triangular areas of AFN and BNS
I.) (5x+3)/4-(2x-4)/3=5
Clear fractions:
3·((5x+3)/4)=15x+9
4·((2x-4)/3)=8x-16
15x+9-(8x-16)=5
15x+9-8x+16=5
Combine like terms:
7x+25=5
7x=-20
x=-20/7
II.) (3/11)·(5/6)-(9/12)·(4/3)+(5/13)·(6/15)
Remember PEMDAS
So first multiply:
3/11·5/6=15/66
9/12·4/3=3/3·1/1=3/3=1
5/13·6/15=1/13·6/3=6/39=2/13
(15/66)-1+(2/13)
Combine:
15/66-1/1=15/66-66/66=-51/66
-51/66+2/3=-51/66+44/66=-7/66
Answer: -7/66 :)
Answer:
28 - 15 + 5 = 18. Add and subtract will get you ,your answer. Bryce has $18
Step-by-step explanation:
Simple math hope this helped
Quadratic formula is x=(-b+-sqrt(b^2-4ac))/(2a)
a= 1, b=-4, and c=3
Therefore, x=(-(-4)+-sqrt((-4)^2-4(1)(3))/(2(1))
Solve for both the plus and minus equations; the answers will be the roots to the quadratic equation.
Direct variation is y=kx and inverse variation is y=k/x, in this case:
y=-16/x
So this is an inverse variation with a constant of -16
