All categories

2 years ago

I WILL GIVE BRAINLIEST Which of the following tables of x– and y–values represents a function? Select all that apply.

Mathematics

1 answer:

Otrada [13]2 years ago

5 0

Answer:

A and C

Step-by-step explanation:

For a table of x and y variables to be a function, each x-value must not have two different y-values assigned or related to it. However, different x-values can have the same y-value.

Therefore, the table in option A and option C are functions because every of their x-value has exactly one y-value that corresponds it.

You might be interested in

Find the area of the square on this centimetre grid

Bess [88]

Answer:

Step-by-step explanation:

the perfect square count as 1

the half square or the triangle count as half

then u got it bro

6 0

2 years ago

Please Help solve this

IgorLugansk [536]

Ok so to find the ratio just do 12/16 = 3/4. that's your common ratio.

The formula for sum of infinite :

a / (1 - r)

a is the first term and r is the ratio

which is

16 / (1 - (3/4))

which is

16 / (1/4)

16 * 4

which is 64

3 0

3 years ago

Maxwell takes 45 minutes to pack 6 boxes. At this rate, how long will he take to Pack 8 boxes?

mestny [16]

1 minute and 4 seconds

6 0

2 years ago

The sum of the digits of a two-digit number is 5. If nine is subtracted from the number, the digits will be reversed. Find the n

Komok [63]

The number is 32
X-9=23

5 0

2 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

2 years ago