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Answer:
y=5x
Step-by-step explanation:
Because it has the same slope, both equations must have 5x as their slope. The 2 equations are not going to have the same y intercept, so the +3 can be ignored and replaced with b. You want to plug in (1,5) into you new equation of y=5x+b, to get 5=5(1)+b. Multiply 5 by 1 to get 5, and then subtract 5 from both sides, which gives you 0=b. This means that the y intercept is the origin, so your new equation is y=5x
Answer:
27
Step-by-step explanation
Sally = 4 x 3 = 12(left) + 15(spent) = 27
Lucy = 24(spent) + 3(left) = 27 (we dont multiply this part bc Lucy spent 9 more dollars than Sally, 3 x 3 is 9 which equals the difference)
Answer:
11 packages
Step-by-step explanation:
24/6=4
44/4=11
Answer:
- It is the last graph: solid line, shaded area over the line x = 2 - x/2
Explanation:
1) <u>Set the algebraigic expression that represents the combinations of sofa and pillow orders:</u>
- Number of sofas: x (given)
- Number of pillows: 2y (given, since they come in pairs)
- Number of items = number of sofas + number of pillows = x + 2y
- Minimum of 4 items in each order (given) ⇒ x + 2y ≥ 4
<u>2) Predict the graph of the inequality x + 2y ≥ 4</u>
- The border line is the equation x + 2y = 4
- You can choose two points to draw a line
- Choose the axis-intercepst:
x = 0 ⇒ 2y = 4 ⇒ y =4/2 ⇒ y = 2 ⇒ point (0,2)
y = 0 ⇒ x = 4 ⇒ point (4,0)
Then the lines goes through (0,2) and (4,0) ... [the four graphs meet this]
- The shading area is above the line because when you solve for y you get y ≥ 2 - x/2, and the line is included because the "equal to" part of the symbol (≥ means greater than or equal to).
- To state that the line is included the graph uses a continous line instead of a dotted one.
<u>3) Conclusion:</u>
That means that the correct graph is the last one: solid line, shaded area over the line y = 2 - x/2.
Note: a more detailed graph would include the fact that the items cannot be negative, i.e. x ≥ 0 and y ≥ 0, which would result in that the shaded area would be on the first quadrant.