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3 years ago

(m+2)(m+3)=(m+2)(m-2) what is the answer please

2 answers:

6 0

The first time you look at this, you would think that you can just cancel the
(m+2) off of each side. But then you're left with (m+3) = (m-2), and there's
no solution for this. So you have to go back and do it the hard way.

Expand each side of the equation. (Clear the parentheses.)

m² + 5m + 6 = m² - 4

Subtract m² from each side:

5m + 6 = -4

Subtract 6 from each side:

5m = -10

Divide each side by 5:

m = -2

Vika [28.1K]3 years ago

5 0

$(m+2)(m+3)=(m+2)(m-2)\\\\m^2+3m+2m+6=m^2-4\\\\m^2-m^2+5m=-4-6\\\\5m=-10\ \ \ \ /:5\\\\m=-2$

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dezoksy [38]

$\sqrt{3}(h-3000)=h$
 $(\sqrt{3}h-\sqrt{3}(3000)=h$
$(\sqrt{3}-1)h=\sqrt{3}(3000)$
$h=\frac{\sqrt{3}(3000)}{(\sqrt{3}-1)}$
$=\frac{\sqrt{3}(3000)}{(\sqrt{3}-1)}$
$=7098.076$ (approximately)

8 0

3 years ago

$7.50 per ticket. the theater earned $450 from ticket sales tuesday's performance. which equation can be used to find the number

Norma-Jean [14]

Answer:

450 ÷ 7.50 = t (tickets sold for Tuesday)

Step-by-step explanation:

this is simple 4th grade division...

4 0

3 years ago

if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?

Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

There are 9 rolls, hence $n = 9$ .

Of the six sides, 2 are 3 or 4, hence $p = \frac{2}{6} = 0.3333$

The desired probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026$

$P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117$

Then:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857$

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0

3 years ago

A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra

VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that $N = 6$

4 blue, which means that $k = 4$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4$

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that $N = 6$

2 yellow, which means that $k = 2$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667$

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that $P(A) = 0.4 + 0.0667 = 0.4667$

Probability of both being the same color and blue:

40% probability that both are blue, which means that $P(A \cap B) = 0.4$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571$

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0

3 years ago

Pythagorean theory m3/9=3

Anna71 [15]

Answer:

Step-by-step explanation:

m3/9=3

We move all terms to the left:

m3/9-(3)=0

We multiply all the terms by the denominator

m3-3*9=0

We add all the numbers together, and all the variables

m^3-27=0

#learnwithbrainly

6 0

2 years ago