Penelope went to the store to buy x pounds of broccoli for $1.10 per pound and y cans of soup for $2.50 each. in total, she spent less than $10. the inequality relating the purchases she made and the total purchase price is
1.10x + 2.50y < 10. which are reasonable solutions for this situation? check all that apply.

(–1, 4)
(0, 2)
(3, 2.5)
(2, 4)
(0.5, 3.78)
(1.5, 3)

(0,2)

(1.5,3)

Step-by-step explanation:

To determine the reasonable solutions for this inequality we simply substitute the values of x and y into the equation and check whether the condition on the right hand side is met. If the condition is met then the set of values are reasonable solutions otherwise if the condition is violated the set of values are not reasonable solutions;

The first alternative cannot be a reasonable solution since the quantity of broccoli cannot be negative.

For the second alternative, we substitute x with 0 and y with 2 in the left hand side;

1.10(0)+2.50(2) = 5. 5 is less than 10 hence (0,2) is a reasonable solution.

For the third alternative, we substitute x with 3 and y with 2.5 in the left hand side;

1.10(3)+2.50(2.5) = 9.55. 9.55 is less than 10 hence (3,2.5) is a reasonable solution but it will not be applicable in this case since we can not purchase a fraction of a can.

For the fourth alternative, we substitute x with 2 and y with 4 in the left hand side;

1.10(2)+2.50(4) = 12.2. 12.2 is greater than 10 hence (2,4) is not a reasonable solution.

For the fifth alternative, we substitute x with 0.5 and y with 3.78 in the left hand side;

1.10(0.5)+2.50(3.78) = 10. 10 is not less than 10 hence (0.5,3.78) is not a reasonable solution.

For the sixth alternative, we substitute x with 1.5 and y with 3 in the left hand side;

1.10(1.5)+2.50(3) = 9.15. 9.15 is less than 10 hence (1.5,3) is a reasonable solution.

7

step-by-step explanation:

the answer would be 7

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.