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3 years ago

F(x) =2x^2 Find F(x+3)

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

6 0

Step-by-step explanation:

it may help you to understand.

You might be interested in

Find the area of the trapezoid

lidiya [134]

area = ( top rule + bottom rule ) × height / 2

area = ( 2 + 3 ) × 4 / 2

area = 5 × 4 / 2

area = 20 / 2

area = 10

4 0

2 years ago

Read 2 more answers

I don’t understand please help!

nikklg [1K]

Answer:

x > 10/3

x will always be greater that 10/3 never the same nor less.

Step-by-step explanation:

3/5x + 1/3 < 4/5x - 1/3

3/5x - 4/5x + 1/3 < -1/3

-1/5x + 1/3 < -1/3

-1/5x < -1/3 - 1/3

-1/5x < -2/3

-1/5x/-1 < -2/3/-1 (Since it is both divided by a negative the sign switches, the sign also switches with multiplication. *Note: it only switches if it is negative!*)

1/5x > 2/3

5 • 1/5x > 2/3 • 5

x > 2/3 • 5/1

x > 10/3

6 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$