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3 years ago

Which of these graphs represents a function

Mathematics

1 answer:

azamat3 years ago

8 0

Answer: C.

Explanation: A function only has one y value for every x value. This means that an x cannot have more than one y value. You can do the vertical line test on graphs to figure out if it is a function or not. In this case, graph C is the only one that passes the vertical line test.

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Solve 2 1/4 + 6% = 6% -4%

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Answer:

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Step-by-step explanation:

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3 years ago

What is the scientific notation for 3800

BigorU [14]

3.8 x 10²

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3 years ago

Work out<br> a) 5 3/4- 2 1/4

xxTIMURxx [149]

Answer:

The answer is 3 1/2.

Step-by-step explanation:

-There are 2 simple ways to do this.

-The first is just subtracting 3/4 and 1/4 which equals 2/4 or 1/2.

-The second way is to turn it into an improper fraction.

-(5*4)+3=23 and (2*4)+1=9.

-23/4-9/4=14/4

-14/4=3 1/2

7 0

2 years ago

Read 2 more answers

5y+4 times 2 what is the coefficient in this expression?

rodikova [14]

Answer: 5

Step-by-step explanation: The coefficient is the number that comes right before the variable. Another way to say this is it's what the variable is multiplied by.

3 0

3 years ago

Suppose we want to choose 5 colors, without replacement, from 8 distinct colors.1. If the order is relevant, how many can be don

Charra [1.4K]

(1) From the information given, if we want to choose 5 colors from 8 distinct colors and the order in which the selection is made is relevant, then what we have is a permutation.

The formula is given as;

$nP_r=\frac{n!}{(n-r)!}$

This formula means we need to select/arrange r items out of a total of n items and the anwer derived would be the total number of arrangements possible.

Therefore, we would have;

$\begin{gathered} nP_r\Rightarrow_8P_5 \\ _8P_5=\frac{8!}{(8-5)!}\Rightarrow\frac{8!}{3!} \\ _8P_5=\frac{8\times7\times6\times\ldots1}{3\times2\times1}\Rightarrow\frac{40320}{6} \\ _8P_5=6720 \end{gathered}$

Therefore, if the order is relevant, this selection can be done in 6,720 ways.

(2) If the order is NOT relevant, then what we need to calculate is a combination and the formula is;

$_nC_r=\frac{n!}{(n-r)!r!}$

The formula can now be applied as follows;

$\begin{gathered} _nC_r\Rightarrow_8C_5 \\ _8C_5=\frac{8!}{(8-5)!\times5!} \\ _8C_5=\frac{8!}{3!\times5!}\Rightarrow\frac{8\times7\times6\times\ldots1}{(3\times2\times1)\times(5\times4\times\ldots1)} \\ _8C_5=\frac{40320}{6\times120} \\ _8C_5=56 \end{gathered}$

If the order is not relevant, then the selection can be done in 56 ways.

3 0

1 year ago