Which numbers are divisible by 5 and 9
1 answer:
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Answer:
Graph A → Equation A → Table B → Property D
Graph D → Equation B → Table A → Property B
Graph B → Equation C → Table C → Property E
Graph C → Equation D → Table E → Property A
Graph E → Equation E → Table D → Property C
Step-by-step explanation:
Graph A corresponds to equation A, x·y = 2, and Table B → Property D
xy = 2 has a vertical asymptote at x = 0, and a horizontal asymptote at y = 0
y = 2/x, when x = 0, y = ∞
Graph D corresponds to equation B, + 1 which corresponds to table A → Property B
The domain extends from (-∞, ∞), and [1, -∞)
Graph B corresponds to equation C, -(x + 2)² which gives the values in Table C and Property E wich is a horizontal shift of 2 units left and a reflection across the x-axis
Graph C corresponds to equation D, (1/3)·x - 2, which gives the values on Table E and Property A
The y-intercept of (1/3)·x - 2 = (0, -2), and the x-intercept = (6, 0)
Graph E corresponds to equation E, x³ = y, which gives the values on Table D Property C, y is x cubed
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=
x=
x=
x=
x=
there are no real numbers that satisfy this
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=
x=
x=
x=
x=
there are no real numbers that satisfy this