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Ilia_Sergeevich [38]
2 years ago
10

Which table represents a function? A 2-column table with 4 rows. The first column is labeled x with entries negative 3, 0, negat

ive 2, 8. The second column is labeled y with entries negative 1, 0, negative 1, 1. A 2-column table with 4 rows. The first column is labeled x with entries negative 5, 0, negative 5, 6. The second column is labeled y with entries negative 5, 0, 5, negative 6. A 2-column table with 4 rows. The first column is labeled x with entries negative 4, negative 2, negative 2, 0. The second column is labeled y with entries 8, 2, 4, 2. A 2-column table with 4 rows. The first column is labeled x with entries negative 4, 3, 1, negative 4. The second column is labeled y with entries 2, 5, 3, 0.
Mathematics
1 answer:
tangare [24]2 years ago
8 0

Answer:

This is a many-to-one function as the value of y = -1 corresponds to two values of x:

\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -3 & -1\\\cline{1-2} 0 & 0\\\cline{1-2} -2 & -1\\\cline{1-2} 8 & 1\\\cline{1-2}\end{array}

Step-by-step explanation:

<u>Function</u>

A special relationship where each input (x-value) has a single output (y-value).

A function is <u>one-to-one</u> if each value in the range (y-values) corresponds to exactly one value in the domain (x-values).

A function is <u>many-to-one</u> if some values in the range (y-values) correspond to more than one (many) value in the domain (x-values).

This is a many-to-one function as the value of y = -1 corresponds to two values of x:

\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -3 & -1\\\cline{1-2} 0 & 0\\\cline{1-2} -2 & -1\\\cline{1-2} 8 & 1\\\cline{1-2}\end{array}

This is not a function as the value of x = -5 corresponds to two values of y:

\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -5 & -5\\\cline{1-2} 0 & 0\\\cline{1-2} -5 & 5\\\cline{1-2} 6 & -6\\\cline{1-2}\end{array}

This is not a function as the value of x = -2 corresponds to two values of y:

\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -4 & 8\\\cline{1-2} -2 & 2\\\cline{1-2} -2 & 4\\\cline{1-2} 0 & 2\\\cline{1-2}\end{array}

This is not a function as the value of x = -4 corresponds to two values of y:

\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -4 & 2\\\cline{1-2} 3 & 5\\\cline{1-2} 1 & 3\\\cline{1-2} -4 & 0\\\cline{1-2}\end{array}

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labwork [276]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

A 12 cm cone with a dome on top of it that has an 8 cm diameter

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cone volume)

V = \dfrac{ \pi *r^2*h}{3}

V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}

V = 3.14*16*4

\boxed{V = 200.96\:cm^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4* \pi * \dfrac{r^3}{3} , but we have a hemisphere, so the formula will be half the volume of the hemisphere V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}

Formula: (Volume of the hemisphere)

V = 2* \pi * \dfrac{r^3}{3}

Solving:

V = 2* \pi * \dfrac{r^3}{3}

V = 2*3.14 * \dfrac{4^3}{3}

V = 2*3.14 * \dfrac{64}{3}

V = \dfrac{401.92}{3}

\boxed{ V_{hemisphere} \approx 133.97\:cm^3}

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

\boxed{\boxed{\boxed{Volume\:of\:the\:figure = 334.93\:cm^3}}}\end{array}}\qquad\quad\checkmark

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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Answer:

I believes the answer is B

Step-by-step explanation:

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