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3 years ago

PLEASE HELP which mapping diagram does NOT represent function from x-y

Mathematics

2 answers:

Tomtit [17]3 years ago

8 0

Diagram 1 I hope this helps

Bas_tet [7]3 years ago

3 0

Answer:

Diagram 1

Step-by-step explanation:

This diagram does not represent a function because the x value 6 is outputting 2 different y values, 4 and 6. This goes directly against the rules of a function.

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17 + 5(x + 1)=37 X= 15 X 3 X=-15 X=-3

aleksklad [387]

Answer:

X=3

Step-by-step explanation:

$17 + 5(x + 1) = 37$

$17 + 5x + 5 = 37$

Group like terms

$5x = 37 - 17 - 5$

$5x = 15$

$\frac{5x}{5} = \frac{15}{5}$

$x = 3$

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3 years ago

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100 divided by what is 30

scZoUnD [109]

Answer:

3.33333333333

Step-by-step explanation: it keeps going on and on.

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3 years ago

What is the blank? Blank+(-6)+(-3)=-17

RideAnS [48]

The blank is -8 because -6 +-8=-14+-3=-17.

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3 years ago

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Extra point hereWhich metric unit would be best to measure the diameter of a pencil?

Roman55 [17]

Answer:

The millimeter.

Step-by-step explanation:

The best choice here is the millimeter (mm).

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4 years ago

In Exercises 1–4, let S be the collection of vectors cxyd in R2that satisfy the given property. In each case, either prove thatS

Artyom0805 [142]

Answer:

S is not the subspace of $R^2$

Step-by-step explanation:

Let us suppose two vectors u and v belong to the S such that the property of xy≥0 is verified than

$u=\left[\begin{array}{c}-1 \\0\end{array}\right]$

$v=\left[\begin{array}{c}0 \\1\end{array}\right]$

Both the vectors satisfy the given condition as follows and belong to the S

$x_u y_u=-1\times 0=0 \geq 0\\x_v y_v=1\times 0=0 \geq 0\\$

Now S will be termed as subspace of R2 if

u+v also satisfy the condition
ku also satisfy the condition

Taking u+v

$u+v=\left[\begin{array}{c}-1 \\0\end{array}\right]+\left[\begin{array}{c}0 \\1\end{array}\right]\\u+v=\left[\begin{array}{c}-1+0 \\0+1\end{array}\right]\\u+v=\left[\begin{array}{c}-1 \\1\end{array}\right]$

Now the condition is tested as

$\\x_{u+v} y_{u+v}=-1\times 1=-1$

This indicates that the condition is not satisfied so S is not the subspace of $R^2$

3 0

3 years ago