2-14:
A) 3x+5y=3(3)+5(2)
=9+10
=19
B) x²+y²=3²+2²
=9+4
=13
C) -5x+9=-5(3)+9
=-15+9
=-6
D) xy-1=(2×3)-1
=6-1
=5
E) 2x²-y=2(3)²-2
=18-2
=16
F) 2y-x+5+x²-3y=2(2)-3+5+3²-3(2)
=4-3+5+9-6
=9
2-15: area=y²+xy+xy
=(y²+2xy) un²
perimeter=y+y+y++y+x+y/2+y+x+y/2
=(6y+2x) un
Answer:
C good luck !!!!
Step-by-step explanation:
blah blah blah
X=-b/2a is the formula for finding the axis of symmetry
So x= -30/2(5)
X=-30/10
X=-3
Because the axis of symmetry is -3, we know where to place our line, and we also know that the parabola is open downwards, which means that the vertex will be maximum. To find the vertex, plug in your values with the axis of symmetry as a midway point. Plug that in for x and so you should have the following:
F(x)
Y(f(x) and y variables are interchangeable) =5(-3)^2-30(-3)+49
Solve for y(f(x))
5(-3)^2-30(-3)+49
(-3)^2=3^2
3^2*5+30*3+49
Multiply
3^2*5+90+49
Add numbers
3^2*5+139
9*5=45
45+139=184
Y=184
So, your vertex would be
(-3,184) and it would be maximum. From there you can plug in the rest of your table of values.
Place the point of the compass on point J and draw an arc in the interior of the triangle.
