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dimaraw [331]
3 years ago
11

Does this graph represent a function? Why or why not?

Mathematics
1 answer:
DedPeter [7]3 years ago
8 0

Answer: C

Step-by-step explanation: For a function, each x-coordinate corresponds to exactly one y-coordinate.

To determine whether the graph shown here

is a function, we can use the vertical line test.

The vertical line test tells us that if each x-coordinate on the graph corresponds to exactly one y-coordinate, then any vertical line that we draw on the graph should hit the graph at only one point.

For the graph show here, any vertical line that you draw with hit the graph at only one point which means it does pass the vertical line test.

So this graph is a <em>function</em>.

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Tossing a regular coin 3 times, what is the probability of getting heads every time? 4)
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The probability of getting heads on the first toss is  1/2 .

Then, the probability of getting heads on the second toss is  1/2 .

Then, the probability of getting heads on the third toss is  1/2 .

The probability of all three things happening is ...

                 (1/2) x (1/2) x (1/2)  =  1/8  =  0.125  =  12.5% .



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Tell whether 4/2 and 14/7 form a proportion
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Find the area of the circle. Round to the nearest tenth.
Lerok [7]

Answer:

283.5

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A=π(9.5^2)

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A rely race is 1/4 mile long and is run by 4-member teams. If each team member runs the same distance, what fraction of a mile d
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3 years ago
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Hiro bought a small carton of milk at lunch. If the approximate dimensions of the milk carton are shown what is the minimum amou
faltersainse [42]

Answer:

C.\ 104\ in^2

Step-by-step explanation:

At first, the question looks like an optimization problem, but since all the dimensions of the carton are given, we only have to compute the total area of the given figure.

Let's calculate the front (and back) areas, which are rectangles

A_1=(6.5)(3)=19.5\ in^2

Now with the lateral rectangles which happen to have the very same dimensions

A_2=19.5\ in^2

Next, we compute the front and back triangles of base 3 in and height 1.5 in

A_3=\frac{1}{2}(3)(1.5)=2.25\ in^2

Now, the lateral inclined rectangles of base 3 in and height 2 in

A_4=(3)(2)=6\ in^2

Finally, the base rectangle who happens to be a square of side 3 in

A_5=(3)(3)= 9\ in^2

This last area, unlike all others, is not doubled because its counterpart is inside the carton and is not part of the lateral area

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A_t=2(19.5)+2(19.5)+2(2.25)+2(6)+9=103.5\ in^2

The closest option to this answer is

C.\ 104\ in^2

5 0
3 years ago
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