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3 years ago

8 over 2x minus 4 over x squared

Mathematics

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

4x-4/x^2

Step-by-step explanation:

So we can first write out the equation.

It would be:

8/2x-4/x^2

First we can get a common denominator:

8/2x*x^2/x^2=8x^2/2x^3

and

4/x^2*2x/2x=8x/2x^3

Now we have

8x^2/2x^3-8x/2x^3

8x^2-8x/2x^3

We can divide everything by x

8x-8/2x^2

divide everything by 2

4x-4/x^2

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Please help with my algebra 1 homework thank you

MrRissso [65]

1) Answer:

$\frac{4m^{10} }{9n^{18} }$

2) Answer:

$8m ^{5}$

The answer for first equation is, option B.

and the Answer for Second equation is option D.

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3 years ago

In the middle school debate club, 30% of the members are in sixth grade. If there are 12 graders in the club, how many total mem

Slav-nsk [51]

The answer is D because you just multiply the percent times 40 to get 12.

6 0

3 years ago

Read 2 more answers

Let m Z1=(3x)º, m 22=(5x-7)", m 23=(4x+15), and m ZAFD=128º. Find m44 and m ZBFD.

olga55 [171]

Answer:

m<4 = 52°

m<BFD = 98°

Step-by-step explanation:

m<1 = (3x)°

m<2 = (5x - 7)°

m<3 = (4x + 15)°

m<AFD = 128°

✔️Find m<4:

m<4 = 180° - m<AFD (angles on a straight line)

Substitute

m<4 = 180° - 128°

m<4 = 52°

✔️m<BFD = m<2 + m<3

Substitute

m<BFD = (5x - 7)° + (4x + 15)°

We need to find the value of x.

Create an equation to find x.

m<1 + m<2 + m<3 = m<AFD (angle addition postulate)

Substitute

3x + 5x - 7 + 4x + 15 = 128°

Add like terms and solve for x

12x + 8 = 128

12x + 8 - 8 = 128 - 8

12x = 120

12x/12 = 120/12

x = 10

m<BFD = (5x - 7)° + (4x + 15)°

Plug in the value of x

m<BFD = 5(10) - 7 + 4(10) + 15

m<BFD = 50 - 7 + 40 + 15

m<BFD = 98°

4 0

3 years ago

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

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3 years ago

Two number cubes are rolled and the product of the two numbers shown is calculated. Calculate P(multiple of 6). Record your answ

Ludmilka [50]

Answer:

1/6

Step-by-step explanation:

5 0

3 years ago