$3\ consecutive\ even\ numbers:\ 2x,2x+2,2x+4\\\\2x(2x+2)+40=(2x+4)^2\\\\
4x^2+4x+40=4x^2+16x+16\\\\
4x^2-4x^2+4x-16x=16-40\\\\-12x=-24\ \ |:(-12)\\\\x=\frac{24}{12}=2\\\\2x=2=4\\\\2x+2=6\\\\2x+4=8\\\\Numbers\ are\ 4,6,8.$

5 0

3 years ago

Please help me!!!!!!!!!!!

shutvik [7]

The given parabola is $y+4=12(x-7)^2$ . Compare with the standard equation of a parabola in vertex form, $y-c=a(x-b)^2$ .

The vertex of this parabola is $(b,c)$ . Compare $y+4=12(x-7)^2$ with the standard equation $y-c=a(x-b)^2$ .

we see $c=-4,b=7$ . Thus, the vertex of the given parabola is $(7,-4)$ .

8 0

4 years ago

Read 2 more answers

Simplify the expression:

krek1111 [17]

Answer:

235a is 1

235b is a^12k+4 i think

3 0

3 years ago