1-Mrs. Thomas wrote the expression below on the board.
2 answers:
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9514 1404 393
Explanation:
Your question covers a good bit of the material in an algebra course. The short answer is, "the same way you solve a numerical equation." The point of algebra is that literals can stand for numbers, and so be manipulated the same way numbers are.
Expressions are evaluated according to the Order of Operations. For equations involving a single variable, the equation specifies what operations are being performed on that variable. To find the vale of the variable (solve for that literal), you need to "undo" the operations that are performed on it. As with many problems that have layers, you work down through the layers from the outside in. Generally, that means working through the list of operations "backwards," undoing the last one first.
<u>Simple example</u>
y = mx + b . . . . . . solve for x
In this equation, the operations performed on x are ...
- multiplication by m
- addition of b to the product
In accordance with the above, the first thing we do is "undo" the addition of b. (Note that this could be a number or literal--or even a complicated expression--and the process would be exactly the same.) To "undo" addition, we add the opposite.
y -b = mx +b -b ⇒ y -b = mx
Next, we "undo" the multiplication by m. That is, we divide by m, or multiply by the reciprocal of m. Either is the same as the other.
(y -b)(1/m) = (mx)(1/m) ⇒ (y -b)/m = x
Now, we have solved this literal equation for x.
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Throughout this process you must adhere strictly to the properties of equality. That is, anything you do to one side of the equation must also be done to the other side.
The reason you study inverses and identity elements is so you understand that addition of an additive inverse produces the additive identity element:
x + (-x) = 0
Similarly, multiplication by the multiplicative inverse (reciprocal) produces the multiplicative identity element.
x · (1/x) = 1
When other operations are involved, such as raising to a power, trig functions, roots, logs, exponentiation, each of these has an associated inverse function that produces an identity:
(x^a)^(1/a) = x^1 = x
arcsin(sin(x)) = x
(√x)^2 = x
10^(log(x)) = x or log(10^x) = x
Some of these inverse functions have restricted domains, so care must be used when solving equations involving them.
When a variable of interest appears on both sides of the equal sign, then you must figure a way to rearrange the equation so the terms with the variable can be combined.
<u>Example</u>:
ax + b = cx +d . . . . . solve for x
ax -cx = d -b . . . . . . subtract (cx+b). (Of course, this is subtracted from both sides of the equation.)
x(a -c) = d -b . . . . . combine x-terms
x = (d -b)/(a -c) . . . . divide by the coefficient of x
Note that we had to divide the entire right-side expression by the x-coefficient, so had to enclose it in parentheses.
<u>More Complicated Example</u>:
A recent Brainly problem asked for the solution to ...
T = 2π√(L/g) . . . . solve for L
Here, L is divided by g, a root taken, and that multiplied by 2π. Undoing these in reverse order, we first divide by 2π, square both sides to undo the root, then multiply by g to undo the division:
The problem posted on Brainly had numbers where some of these variables are. That does not affect the solution method, except that sometimes numerical values can be combined where literal values cannot.
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<u>Key Points</u>
- The equal sign is sacred, and its truth must be preserved at every step.
- Literal equations are solved the same way numerical equations are solved.
- Inverse operations and functions are used to "undo" operations and functions.
- The Order of Operations can be helpful when considering what to do first.
Answer:
Step-by-step explanation:
In order to answer all of these questions we need the position function and the acceleration function. We will discuss why when we get there.
The velocity function is given; in order to find the position function we have to take the antiderivative of the velocity function. So in order are the position, velocity, and acceleration functions below:
I know the constant on the position function is 1 because the info given tells me that s(0) = 1.
For the first question, the formula to find the average velocity is as follows:
To find s(t2) and s(t1) we sub in 8 for t2 and 0 for t1 to get the following:
That simplifies to
The second question wants the instantaneous velocity at t = 5. We get this by subbing in a 5 for t in the velocity function:
and
v(5) = -2. This means that the velocity of the particle is 2 m/sec, but it is now going in the opposite (or negative) direction.
The third question is asking for the time interval when the particle is moving to the right. On a velocity/time graph, the x's represent the time and the y's represent the velocity. If the "y" values are positive, then the velocity is positive and that means the object is moving to the right. Where the "y" values are negative, that means that the velocity is negative and the object is moving to the left. To find the answer to this problem we think about the positive and negative y values. Because this is a parabola, I know that the places where the graph goes through the x-axis is where the velocity changes from positive to negative and back to positive. In order to find those places where the graph goes through the x-axis I have to factor the velocity function. When I throw that into the quadratic formula on my calculator I get that x = 3 and x = 6. Completing the square on the velocity function gives me a vertex of
Because the y value of the vertex is negative, that means that the values of x from negative infinity to x = 3 give positive velocity values, between x = 3 and x = 6 the values of the velocity are negative, and from x = 6 to x = positive infinity the velocity is positive again.
So to sum up question 3, the particle is moving to the right on the intervals
(-∞, 3] and [6, ∞)
The fourth question is really tricky. It requires you to remember some of your Physics and how velocity and acceleration vectors are related. If the acceleration and the velocity both have the same sign, whether it be positive or negative, the object is speeding up. If the acceleration and velocity vectors have opposite signs, one positive and one negative, then the object is slowing down. We already know that from negative infinity to 3 the velocity is positive, so let's check the acceleration values in that interval. I only need to test one number, so let's test a(2).
a(2) = 2(2) - 9 and a(2) = -5
-5 means the object is slowing down here since the velocity is positive and the acceleration is negative.
Let's test the interval between 3 and 6. Let's test a(4).
a(4) = -1
Between the interval of 3 and 6 seconds, the velocity is negative. Since the acceleration is also negative, the object is speeding up between the time interval [3, 6].
We already found that from the left of t = 3, the object was slowing down; so it would also be slowing down to the right of t = 6.
Phew! That's it! We're done!