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2 years ago

At which value in the domain does f(x) = 0?

Mathematics

1 answer:

bekas [8.4K]2 years ago

5 0

Answer:

Step-by-step explanation:

f(x)=0 at points of intersection of the graph of the function with x-axis.

From the diagram you can see that the graph of the function intersects x-axis at three points:

somewhere between -3 and -2;

somewhere between -1 and 0;

at x=1.

Since x=1 is the value of x from the domain, then correct choice is option C.

You might be interested in

The library has DVDs on crime, adventure and science fiction.

kozerog [31]

Answer:

The library has 1,056 crime DVDs.

Step-by-step explanatiion:

Crime : Adventure : Science

11x : 6x : 10x

6x+384=10x

384=4x

x=96

The number of crime DVDs is 11x.

We found that x=96.

So, the number of crime DVDs the library has is equal to 11*96.

11*96=1,056.

Hope this helps!

6 0

2 years ago

Please help me! And can you please show how you found the answer?

Pani-rosa [81]

Answer:

what is the question I didn't get it please explain what we have to find

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Describe how the graph of the function g(x) = -5x²-2 is related to the graph of

zhenek [66]

Answer:

3 0

1 year 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

What is 47.2 % of 300 ?

kupik [55]

For this type of question, you just need to multiply the percentage with the number
so, the calculation would be

47.2% x 300

47.2/100 x 300

47.2 x 3

= 141.6

6 0

3 years ago

Read 2 more answers