Find the area; reduce to lowest terms:
2 answers:
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Answer:
Nora received <u>35</u> purple pins.
Step-by-step explanation:
If Nora received 50 pins, and 70% of the pins were purple, to calculate the number of purple pins Nora received, we must find 70% of 50.
To find 70% of 50, we can multiply 50 by the decimal form of 70%:
Therefore, we can conclude that since 70% of the 50 pins Nora received were purple, and 70% of 50 is 35, Nora received 35 purple pins.
<h2>Hello!</h2>
The answer is:
The point that is a solution to the system of inequalities is C(4,2).
<h2>Why?</h2>To find the point that is a solution to the system of inequalities, we need to evaluate it into the given inequalities. If the point is a solution to the system of inequalities, both inequalities will be satisfied.
We are given the inequalities:
So, substituting the given points into the given inequalities, we have:
- A. (2,1)
Substituting into the first inequality, we have:
Substituting into the second inequality, we have:
Therefore, since 1 is not less or equal to -2, the point A(2,1) is not a solution to the system of inequalities.
- B. (-2,-1)
Substituting into the first inequality, we have:
Substituting into the second inequality, we have:
Therefore, since -1 is not less or equal to -6, the point B(-2,-1) is not a solution to the system of inequalities.
- C. (4,2)
Substituting into the first inequality, we have:
Substituting into the second inequality, we have:
Therefore, since 2 is less than 6, and 2 is less than 4, the point B(4,2) is a solution to the system of inequalities.
- D(1,3)
Substituting into the first inequality, we have:
Substituting into the second inequality, we have:
Therefore, since 3 is not less than 0, and 3 is not less than -2, the point D(1.3) is not a solution to the system of inequalities.
Hence, the point that is a solution to the system of inequalities is C(4,2).
Have a nice day!
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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.
_____
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.
_____
<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.