Plug the x-values from the table into the given equation and solve for y.
y = 5 - x²
<u>First column</u>
replace x with -2
y = 5 - (-2)²
= 5 - (4)
= 1
(x, y) = (-2, 1)
<u>Second column</u>
replace x with -1
y = 5 - (-1)²
= 5 - (1)
= 4
(x, y) = (-1, 4)
<u>Third column</u>
replace x with 0
y = 5 - (0)²
= 5 - (0)
= 5
(x, y) = (0, 5)
<u>Fourth column</u>
replace x with 1
y = 5 - (1)²
= 5 - (1)
= 4
(x, y) = (1, 4)
<u>Fifth column</u>
replace x with 2
y = 5 - (2)²
= 5 - (4)
= 1
(x, y) = (2, 1)
Answer:
1/24
Step-by-step explanation:
13/6x = 52
1/6x = 52/13
1/6x = 4
1 = 4(6x)
1 = 24x
24x = 1
x = 1/24
<u>M</u><u>e</u><u>t</u><u>h</u><u>o</u><u>d</u><u> </u><u>1</u><u> </u><u>:</u>
replace x and y by their value 1 and 3
3 = 4(1) - 1 = 4-1 = 3
2(1) + 3 = 2 + 3 = 5
correct
<u>M</u><u>e</u><u>t</u><u>h</u><u>o</u><u>d</u><u> </u><u>2</u><u> </u><u>:</u>
y = 4x - 1
2x + y = 5
y = 4x - 1
y = -2x + 5
y - y = 4x - 1 - ( -2x + 5 )
0 = 4x - 1 + 2x - 5
6x - 6 = 0
6x = 6
x = 6/6 = 1
y = 4x - 1
y = 4(1) - 1
y = 3
correct
First you would distribute the 2 to the -8 (because of the parenthithese), which would give you -16.
Then you would add the -3 .
So, your answer would be -19. Hope this helps!
When you have an equation that has variables and you know the value of those variables, just plug it it. So plugging the vaules we get
Just evaluating we get 4
