The quotient of a number and 3 decreased by 121 is -164. Select the solution to the sentence shown.
1 answer:
Its algebra. The original equation is 
To solve for a variable, we reverse the order of operations, beginning with addition/subtraction, and then multiplication/division. To remove a number from one side, we must do the opposite to the other side. In this case, to get rid of the -121 we must add 121 to the -164. This gives us -43. Then, to get the x by itself, we must multiply the other side by 3. -43*3=129
When we are doing the opposite of an operation to the other side, we are really reversing the operation and, to keep both sides equal, we must do whatever we have done to one side to the other side. So when we have -121, we add 121 as it equals 0, therefore it is gone. Since a equation must be balanced, we have to do what we did to the other side (adding 121).
To solve for a variable, we reverse the order of operations, beginning with addition/subtraction, and then multiplication/division. To remove a number from one side, we must do the opposite to the other side. In this case, to get rid of the -121 we must add 121 to the -164. This gives us -43. Then, to get the x by itself, we must multiply the other side by 3. -43*3=129
When we are doing the opposite of an operation to the other side, we are really reversing the operation and, to keep both sides equal, we must do whatever we have done to one side to the other side. So when we have -121, we add 121 as it equals 0, therefore it is gone. Since a equation must be balanced, we have to do what we did to the other side (adding 121).
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Answer:
When we have a rational function like:
The domain will be the set of all real numbers, such that the denominator is different than zero.
So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.
Then we need to solve:
x^2 + 3 = 0
x^2 = -3
x = √(-3)
This is the square root of a negative number, then this is a complex number.
This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.
D: x ∈ R.
b) we want to find two different numbers x such that:
r(x) = 1/4
Then we need to solve:
We can multiply both sides by (x^2 + 3)
Now we can multiply both sides by 4:
Now we only need to solve the quadratic equation:
x^2 + 3 - 4*x - 4 = 0
x^2 - 4*x - 1 = 0
We can use the Bhaskara's formula to solve this, remember that for an equation like:
a*x^2 + b*x + c = 0
the solutions are:
here we have:
a = 1
b = -4
c = -1
Then in this case the solutions are:
x = (4 + 4.47)/2 = 4.235
x = (4 - 4.47)/2 = -0.235