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3 years ago

Which of these relations are functions?

Mathematics

1 answer:

Brums [2.3K]3 years ago

3 0

Answer: None of them are functions.

Step-by-step explanation:

By definition, a relation is a function if and only if each input value have an unique output value.

You need to remember that the input values are the x-values and the output values are the y-values.

In the first table you can observe that the input value 2 has two different output values:

$x=2;y=11\\\\x=2;y=15$

Therefore, this relation is not a function.

Notice that in the second table the input value -4 has two different output values:

$x=-4;y=-2\\\\x=-4;y=-5$

Therefore, this relation is not a function.

In the set $\{(-5,-7), (-2,-7), (7,17), (-5,21)\}$ , observe that the input value -5 has two different output values:

$x=-5;y=-7\\\\x=-5;y=21$

Therefore, this relation is not a function.

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If 50% of a number is 300, what is the numbers?

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50 per cent can also be termed as a half of the number.

therefore 300 is a ½ of the original number

300 multiplied by 2

answer= 600

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4 years ago

Plz help ill give you brainlist

Nana76 [90]

1=3
2=4
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3 years ago

A gardening club has 21 members, of which 13 are males and the rest are females. What is the ratio of females to all

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Answer:

There are 8 Females out of the 21

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3 years ago

Jim's monthly salary is $925. What is his annual salary?

Travka [436]

Answer:

11,100

Step-by-step explanation:

There's 12 months in a year. Annual is yearly. So, the equation may look like this:

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8 0

3 years ago

Read 2 more answers

Starting at an initial height of 364 feet, an object is projected

Lana71 [14]

Step-by-step explanation:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

v_0 = 144 ft/sv

=144ft/s

The equation that describes its height at time t is

s(t) = -16t^2 + 144 ts(t)=−16t

+144t (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

v(t) = s'(t) = -32t +144v(t)=s

′

(t)=−32t+144 (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

\begin{gathered}0=-32t+144\\t=\frac{144}{32}=4.5 s\end{gathered}

0=−32t+144

144

=4.5s

And by substituting this value into eq.(1), we also find the maximum height:

s(t) = -16(4.5)^2+144(4.5)=324 fts(t)=−16(4.5)

+144(4.5)=324ft

<h3>used this way i even don't know sorry:))</h3>

7 0

2 years ago