All categories

3 years ago

An equation is shown below:

2 answers:

7 0

Part A: first multiply 2 by x and -3, then combine the like terms, after that add 6 from both sides, then subtract 6x from both side, the answer is -5.
4x + 2(x-3) = 4x + 2x - 11
4x +2x -6 = 4x + 2x - 11
6x -6 = 6x - 11
6x = 6x -11 +6
6x = 6x -5
6x-6x= -5
0 = -5
part B) add the like terms

Eva8 [605]3 years ago

5 0

Answer: There is no solution to the equation ad we have used distributive property to solve the equation.

Step-by-step explanation: The given equation is

$4x+2(x-3)=4x+2x-11.$

PART A:

The solution of the above equation is as follows:

$4x+2(x-3)=4x+2x-11\\\\\Rightarrow 4x+2x-6=6x-11\\\\\Rightarrow 6x-6=6x-11\\\\\Rightarrow -6=-11,$

which cannot be TRUE.

So, the equation will not have any solution.

PART B: We have used DISTRIBUTIVE PROPERTY to solve the equation, which states that

for any real numbers a, b and c

(i) a(b + c) = ab + ac,

(ii) (a + b)c = ac + bc.

Thus, there is no solution to the equation ad we have used distributive property to solve the equation.

You might be interested in

4x-9y=9

Alexxandr [17]

Multiply both sides of the second equation by 4. That will give you -4x in the second equation which when added to 4x of the first equation will eliminate x.

Second equation:
-x + 3y = 6

Multiply the second equation by 4 on both sides:
-4x + 12y = 26

5 0

3 years ago

Math Math Math Math help help help

Alchen [17]

Answer: 5a: 5/8

Step-by-step explanation: sorry, thats the only one i can read `:)

5 0

3 years ago

Read 2 more answers

Use a calculator to find the missing side length. Tell whether the side lengths form a Pythagorean Triple.

12345 [234]

Pythagorean theorem: a^2 + b^2 = c^2
The hypotenuse is always the c in the equation.
If the two side lengths they gave you are not the hypotenuse, then enter
4.5^2 + 5.6^2 in the calculator, then find the square root of that. :-)

5 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

Course Activity: Relationships Betwee

Alenkinab [10]

Answer:

yes

Step-by-step explanation:

A rational number is in the form p/q where p and q are both integers and the denominator q is not equal to zero. This means that the set of all real numbers are rational numbers.

Since q can be any integer apart from zero, it means that all integers are rational numbers i.e for q = 1.

Hence 2 which can be written as fraction as 2/1 is a rational number since the denominator (q = 1) is a non zero number

5 0

3 years ago