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3 years ago

Louise begins factoring the polynomial, which has four terms.

Mathematics

1 answer:

ad-work [718]3 years ago

5 0

Louise wants to factorize completely the given polynomial $4x^{3}+12x^{2}+5x+15$

Grouping first, second terms together and third, fourth terms together

= $4x^{3}+12x^{2}+5x+15$

Taking $4x^{2}$ common from first and second terms of the given expression and taking 5 common from the third and fourth terms.

= $=4x^{2}(x+3)+5(x+3)$

Taking (x+3) common from the given expression,

= $=(4x^{2}+5)(x+3)$ is the completely factored form.

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A computer can be classified as either cutting dash edge or ancient. Suppose that 86% of computers are classified as ancient.

Leno4ka [110]

Answer:

(a) The probability that both computers are ancient is 0.7396

(b) The probability that all seven computers are ancient is 0.3479

(c) The probability that at least one of seven randomly selected computers is cutting dash edge is 0.6520.

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

Step-by-step explanation:

We know that 86% of computers are classified as ancient. This means, if one computer is chosen at random, there is an 86% chance that it will be classified as ancient.

$P(ancient)=0.86$

(a) To find the probability that two computers are chosen at random and both are ancient you must,

The probability that the first computer is ancient is $P(ancient)=0.86$ and the probability that the second computer is ancient is $P(ancient)=0.86$

These events are independent; the selection of one computer does not affect the selection of another computer.

When calculating the probability that multiple independent events will all occur, the probabilities are multiplied, this is known as the rule of product.

Let A be the event "the first computer is ancient" and B the event "the second computer is ancient".

$P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396$

(b) To find the probability that seven computers are chosen at random and all are ancient you must,

Following the same logic in part (a) we have

Let A be the event "the first computer is ancient",

B the event "the second computer is ancient",

C the event "the third computer is ancient",

D the event "the fourth computer is ancient",

E the event "the fifth computer is ancient",

F the event "the sixth computer is ancient", and

G the event "the seventh computer is ancient"

$P(A\:and \:B\:and \:C\:and \:D\:and \:E\:and \:F\:and \:G)=\\P(A)\cdot P(B)\cdot P(C)\cdot P(D)\cdot P(E)\cdot P(F)\cdot P(G) =(0.86)^7=0.3479$

(c) To find the probability that at least one of seven randomly selected computers is cutting dash edge you must

Use the concept of complement. The complement of an event is the subset of outcomes in the sample space that are not in the event.

Let C the event "the computer is cutting dash edge".

Let A the event "the seven computers are ancient".

$P(C)=1-P(A)=1-0.3479=0.6520$

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

5 0

3 years ago

In her first 20 games, Jennie served 4 aces. If she continues to serve aces at this rate, how many aces will she have after 160

Talja [164]

Answer:

32 aces

Step-by-step explanation:

4/20= x/160

once you set up this proportion you can cross multiple to get:

20x= 640

then just solve for x:

x= 32

5 0

3 years ago

Which of the following measurements is a little more than a yard? A. Centiliter B. 1/4 foot C. Centimeter D. Meter

BabaBlast [244]

D? Maybe I think I’m right hopefully!!

3 0

3 years ago

Can some answer pls.

kirill [66]

Answer:

E. (7,-2)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

. Quantas senhas com 4 algarismos diferentes podemos escrever com os algarismos 1, 2, 3, 4, 5, 6?

ruslelena [56]

Answer:

We can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

Step-by-step explanation:

We have to find how many passwords with 4 different digits can we write with the numbers 1, 2, 3, 4, 5, and 6.

Firstly, it must be known here that to calculate the above situation we have to use Permutation and not combination because here the order of the numbers in a password matter.

Since we are given six numbers (1, 2, 3, 4, 5, and 6) and have to make 4 different digits passwords.

Now, for first digit of the password, we have 6 possibilities (numbers from 1 to 6).

Similarly, for second digit of the password, we have 5 possibilities (because one number from 1 to 6 has been used above and it can't be repeated).

Similarly, for the third digit of the password, we have 4 possibilities (because two numbers from 1 to 6 have been used above and they can't be repeated).

Similarly, for the fourth digit of the password, we have 3 possibilities (because three numbers from 1 to 6 have been used above and they can't be repeated).

So, the number of passwords with 4 different digits we can write = $6 \times 5 \times 4 \times 3$ = 360 possibilities.

Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

4 0

3 years ago