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Softa [21]
2 years ago
8

A graph titled Cost of a Lunch in a School's Cafeteria has Days since school started on the x-axis and cost of lunch (dollars) o

n the y-axis. A horizontal line is at y = 2.5.
Which is correct about the graph’s slope?
Its slope is positive.
Its slope is negative.
Its slope is zero.
It has no slope.
Mathematics
1 answer:
denpristay [2]2 years ago
3 0

The true statement about the graph's slope is (c) Its slope is zero.

<h3>How to determine the true statement?</h3>

From the question, we have:

A horizontal line is at y = 2.5.

The horizontal line implies that the cost of lunch do not change

Horizontal lines have a slope of 0

Hence, the true statement about the graph's slope is (c) Its slope is zero.

Read more about slope at:

brainly.com/question/3493733

#SPJ1

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4 years ago
EXTRA POINTS
hram777 [196]

Answer:

Step-by-step explanation:

The volume of the pyramid = (1/3)*area of base *height

= (1/3)*10*24*13 = 1040 cubic units.

The total surface area = area of rectangular base + area of 2 isosceles triangles with a base of 24 units + area of 2 isosceles triangles with a base of 10 units.

Area of rectangular base = 24*10 = 240 sq units.

The slant height of isosceles triangles with a base of 24 units = [(10/2)^2+13^2]^0.5 = [25+169]^0.5 = 194^0.5 = 13.92838828 units.

The area of 2 isosceles triangles with a base of 24 units 2*24*13.92838828/2 = 334.2813187 sq units.

The slant height of isosceles triangles with a base of 10 units = [(24/2)^2+13^2]^0.5 = [144+169]^0.5 = 194^0.5 = 17.69180601 units.

The area of 2 isosceles triangles with a base of 10 units 2*10*17.69180601/2 = 176.9180601 sq units.

The total surface area of the pyramid = 240 + 334.2813187 + 176.9180601 = 591.9731247 sq units.

8 0
3 years ago
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Answer:

  r = 120t/(120 -t)

Step-by-step explanation:

Given t = f(r), solve for r.

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Answer:

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Step-by-step explanation:

You want factors of 2 that sum to -3. The positive sign on +2 tells you they both need to have the same sign. The only two negative integer factors of 2 do that.

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