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3 years ago

Classify the number 1/2 into as many categories as it belongs: Natural number, whole number, integer, or rational number.

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

6 0

Answer:

1/2 is a rational number.

Step-by-step explanation:

A rational number can be expressed as a ratio of two numbers. 1/2 is a ratio.

It is also a real number, however you do not list that as a category, so from what you listed, it is only a rational number.

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45001.05

Step-by-step explanation:

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2 years ago

A baseball is thrown into the air from the top of a 224-foot tall building. The baseball's approximate height over time can be r

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The function is $h(t) = -16t^2 + 80t + 224$ , and according to the description of the function in the problem statement, we have the following:

at t=0 after being thrown (that is, at initial time), the height of the ball is calculated by h(0) as follows:

$h(0) = -16(0)^2 + 80(0) + 224=0+0+224=224$ (ft), which is the initial height, as expected.

At t=1 (sec), the height would be $h(1) = -16(1)^2 + 80(1) + 224=-16+80+224=288$ .

etc.

The path is parabolic, as we know by seeing that the function is a quadratic polynomial function. This function has been given in factored form as well. From that we can see that the zeros of the function are t=7 and t=-2.

This means that at t=7 sec, the height h is 0, which means that the ball has hit the ground. t=-2 has no significance in the context of our problem so we just neglect it.

Answer: B) 7 sec

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3 years ago

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What percent of a dollar is a nickel and a penny?<br><br><br><br> It is ______% of a dollar.

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Answer:

6 %

Step-by-step explana

100 pennies makes a dollar... you just add 5 for the nickel and 1 fo the penny and it makes 6

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3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

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3 years ago

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Hii!

So I did this and I have A!

A: You cannot get the mean from the graph but you CAN get the third quartile!

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C: An outlier would be much larger than the rest of the data or much smaller than the rest of the data. An outlier would make the "whisker" portion longer and could potentially slightly shift the box.

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3 years ago

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