A school typically sells 500 yearbooks each year for $50 each.
The economics class does a project and discovers that they can sell 100 more yearbooks for every $5 decrease in price.The revenue for yearbook sales is equal to the number of yearbooks sold times the price of the yearbook.

Let X represent the number of $5 decreases in price. If the expression that represents the revenue is written in the form R(X)=(500+ax)(50-bx). Find the values of a and b.

Zeros of the given equation [tex]3x^{2} -7x+ 1[/tex] are [tex]\frac{7+\sqrt{37} }{6} \ or \frac{7-\sqrt{37} }{6}[/tex].

The zeros of a quadratic equation f(x) are all the x-values that make the polynomial equal to zero.

The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula:

[tex]x = \frac{-b \pm \sqrt{b^{2} -4ac} }{2a}[/tex]

According to the given question.

We have a function.

[tex]f(x) = 3x^{2} -7x+1[/tex]

To find the zeros of the function equate f(x) = 0.

[tex]3x^{2} -7x+1 = 0\\[/tex]

Solve the above equation by quadratic method.

[tex]x = \frac{7\pm\sqrt{(7)^{2} -4(3)(1)} }{2(3)}[/tex]

[tex]\implies x = \frac{7\pm\sqrt{49-12} }{6}[/tex]

[tex]\implies x = \frac{7\pm\sqrt{37} }{6}[/tex]

[tex]\implies x = \frac{7+\sqrt{37} }{6} \ or \frac{7-\sqrt{37} }{6}[/tex]

Hence, zeros of the given equation [tex]3x^{2} -7x+ 1[/tex] are [tex]\frac{7+\sqrt{37} }{6} \ or \frac{7-\sqrt{37} }{6}[/tex].

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we have  

[tex]0.5-\left|x-12\right|=-0.25[/tex]  

we know that        

The absolute value has two solutions

Subtract  [tex]0.5[/tex] both sides

[tex]-\left|x-12\right|=-0.25-0.5[/tex]  

[tex]-\left|x-12\right|=-0.75[/tex]  

Step 1

Find the first solution (Case positive)

[tex]-[+(x-12)]=-0.75[/tex]

[tex]-x+12=-0.75[/tex]

Subtract  [tex]12[/tex] both sides

[tex]-x+12-12=-0.75-12[/tex]

[tex]-x=-12.75[/tex]

Multiply by [tex]-1[/tex] both sides

[tex]x=12.75[/tex]

Step 2

Find the second solution (Case negative)

[tex]-[-(x-12)]=-0.75[/tex]

[tex]x-12=-0.75[/tex]

Adds  [tex]12[/tex] both sides

[tex]x=-0.75+12[/tex]

[tex]x=11.25[/tex]

Statements

case A) The equation will have no solutions

The statement is False

Because the equation has two solutions------> See the procedure

case B) A good first step for solving the equation is to subtract 0.5 from both sides of the equation

The statement is True ----->  See the procedure

case C) A good first step for solving the equation is to split it into a positive case and a negative case

The statement is False ----->  See the procedure

case D) The positive case of this equation is 0.5 – |x – 12| = 0.25

The statement is False

Because the positive case is [tex]0.5-(x-12)=-0.25[/tex] -----> see the procedure

case E) The negative case of this equation is x – 12 = –0.75

The statement is True -----> see the procedure

case F) The equation will have only 1 solution

The statement is False

Because The equation has two solutions------> See the procedure

The equation 0.5 - |x - 12| = -0.25 has no solutions because an absolute value cannot be negative. Attempting to split the equation into positive and negative cases or solving for x is fruitless because the left side of the equation will always be at least 0.5.

When solving the equation 0.5 - |x - 12| = -0.25, we can immediately notice that it will have no solutions because the absolute value is always non-negative, and therefore the left-hand side cannot be less than 0.5. Hence, subtracting 0.5 from both sides is not a good first step. Instead, you would typically isolate the absolute value on one side, but given that the equation equals a negative number, we know it has no solutions without additional steps.

Additionally, splitting the equation into a positive case and a negative case isn't useful here, because no matter what's inside the absolute value, the output cannot lead to a negative result, thus making both cases moot.

The statements that say "The positive case of this equation is 0.5 - |x - 12| = 0.25" and "The negative case of this equation is x - 12 = -0.75" are incorrect as they misinterpret how the absolute value works. Lastly, the equation does not have any solution, so it cannot have only one solution.

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The expression that satisfies the given condition is using if-else conditional statements to check the value of the variable 'month' and assign the corresponding month name.

The expression that satisfies the given condition is:

if (month == 1) {     answer = "jan"; } else if (month == 2) {     answer = "feb"; } else if (month == 3) {     answer = "mar"; } // ... continue this pattern for the remaining months

This code uses conditional statements (if-else) to check the value of the variable 'month' and assigns the corresponding month name to the variable 'answer'.

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They can differ only by a constant is true of f(x) and g(x) if both are antiderivatives of f(x) Hence, option D is correct.

When two functions, F(x)  and G(x), are antiderivatives of the same function f(x), it means that their derivatives are equal to f(x).

This relationship can be represented as:

F'(x) = G'(x) = f(x)

However, it's important to note that if  F(x) and G(x) are both antiderivatives of F(x), then their difference, F(x) - G(x), will have a derivative of zero.

Consequently, F(x) and G(x) can differ only by a constant.

So, the correct option is D.

Complete question:

What must be true of f(x) and G(x) if both of them are antiderivatives of f(x)?

A. They are the same function

B. They can differ by a factor of x

C. If is not possible for two functions to be antiderivatives of the same function

D. They can differ only by a constant

Answer: [tex]x-5=\pm 3[/tex] will be the next step of the given expression.

Step-by-step explanation:

Since, Given expression is |x-5|+2=5

On solving the above expression,

Step 1.  [tex]|x-5|+2=5[/tex]

step 2. [tex]|x-5| = 5-2[/tex]

Step 3. [tex]|x-5| = 3[/tex]

Step 4. [tex]x-5=\pm 3[/tex] (because mode takes both positive and negative values)

Answer:

-20/29

Step-by-step explanation:

The question concerns a basic algebraic expression 'eight times the sum of a and b' which is represented as 8(a + b). The expression emphasizes the operation order: sum first, then multiply, which results in a value eight times greater than the original sum.

The question 'eight times the sum of a and b' is a mathematical expression that can be represented as 8(a + b). This expression means you first add the numbers 'a' and 'b' and then multiply their sum by eight. The result you get after the multiplication will be eight times greater than the original sum of 'a' and 'b'. For instance, if 'a' is 2 and 'b' is 3, their sum is 5, and when this is multiplied by eight, it becomes 40, which is the desired expression's value.

To understand this concept further, we can refer to the exponentiation rule mentioned, which states that (xa)b = xa.b. Although this is a different type of operation—exponentiation—it demonstrates a similar principle of first performing the operation inside the parentheses and then applying the outside operation.

Finally, when performing algebraic operations, it's essential to remember that whatever you do to one side of the equation, you should do to the other side to maintain balance. This is a fundamental principle in algebra that helps to solve equations, such as the example provided showing how to isolate 'a' by subtracting 'x' from both sides of the equation a b.

Answer:

5.2

Step-by-step explanation:

The hypothesis in the given mathematical conditional statement 'If the sum of two angles is 90°, then the angles are complementary.' is 'the sum of two angles is 90°'.

In a conditional statement in mathematics, the 'if' part of the statement is called the hypothesis and the 'then' part is termed the conclusion. Given the statement 'If the sum of two angles is 90°, then the angles are complementary.', the hypothesis of this statement is 'the sum of two angles is 90°'.

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In a conditional statement, the hypothesis is the condition that needs to be met. In this case, the hypothesis of the statement 'If the sum of two angles is 90°, then the angles are complementary,' is 'the sum of two angles is 90°.'

In the context of the given conditional statement, 'If the sum of two angles is 90°, then the angles are complementary,' the hypothesis refers to the clause immediately after 'if.' This indicates the condition that needs to be fulfilled for the conclusion to be considered valid. Therefore, the hypothesis for this statement is 'the sum of two angles is 90°'.

After the 'if,' the hypothesis is given, and after the 'then,' you find the conclusion. The conclusion in this case is 'the angles are complementary.'

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A = hours for plan A
B = hours for plan B

Monday: 6A + 5B = 7
Tuesday: 2A + 3B = 3

use elimination by multiplying the 2nd equation by 3.

Doing that we get 3(2A + 3B = 3) = 6A + 9B = 9

So the two equations are now:
6A + 9B = 9

6A + 5B = 7

Subtract and we have 4B = 2

B = 2/4 = 1/2 of an hour

Now put 1/2 back into either equation to solve for A

6A + 5(1/2) = 7
6A + 5/2 = 7
6A = 14/2 -5/2
6A = 9/2
divide by 6 to get A = 9/12 = ¾  hours

Plan A = 3/4 hour

 Plan B = 1/2 hour

Final answer:

By setting up and solving a system of equations, we find that Plan A lasts for 45 minutes per session and Plan B lasts for 30 minutes per session.

Explanation:

Solving for the Duration of Workout Plans

We have information regarding the total duration of workouts and the number of clients for two consecutive days. To find the duration of each workout plan, we use a system of equations. Let A represent the duration of Plan A and B represent the duration of Plan B. The equations based on the given information are:

6A + 5B = 420 minutes (7 hours on Monday)

2A + 3B = 180 minutes (3 hours on Tuesday)

Multiplying the second equation by 3 gives us:

6A + 9B = 540

Subtracting the first equation from this result gives us:

4B = 120 minutes, therefore, B = 30 minutes

Now we substitute B = 30 in the first equation:

6A + 150 = 420, which simplifies to 6A = 270, hence A = 45 minutes

Thus, Plan A lasts for 45 minutes and Plan B lasts for 30 minutes.

An underlined greater than sign in mathematics represents a strict inequality, indicating that one value is significantly greater than another.

In mathematics, an underlined greater than sign usually represents an inequality. When a greater than sign (>) is underlined, it indicates a strict inequality, meaning that the value on the left side is significantly greater than the value on the right side.

For example, if we have the underlined inequality 5 > 3, it means that 5 is larger than 3 and there is a clear distinction between the two values.

It's important to note that this is just one possible interpretation of an underlined greater than sign, as the context in which it is used can vary.

Whole numbers are sometimes integers. Correct option is b.

Integers include both positive and negative whole numbers, as well as zero. Whole numbers are a subset of integers, but they do not include negative numbers. So, while all whole numbers are integers, not all integers are whole numbers.

Relationship between Whole Numbers and Integers:

Every whole number is an integer: Since whole numbers include zero and all positive counting numbers, they are also part of the set of integers.

Not every integer is a whole number: Integers also include negative numbers, which are not part of the set of whole numbers.

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There are 243 ways to select five elements in order from a set of three elements with repetition allowed.

When selecting five elements in order from a set with three elements and repetition is allowed, each selection can include any of the three elements, repeated as necessary. Here's the breakdown:

1. For the first position, there are 3 choices.

2. For the second position, there are also 3 choices, as repetition is allowed.

3. Similarly, for the third, fourth, and fifth positions, there are 3 choices each.

To find the total number of ways, we multiply the number of choices for each position:

3 choices for the first position × 3 choices for the second position × 3 choices for the third position × 3 choices for the fourth position × 3 choices for the fifth position = [tex]\(3^5 = 243\)[/tex] ways.

Therefore, there are 243 different ways to select five elements in order from a set with three elements when repetition is allowed.

The correct naswer is 21.

The number of ways to select five elements in order from a set with three elements when repetition is allowed can be represented in LaTeX as:

[tex]\binom{5+3-1}{5} = \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = 21[/tex]

Explanation:

- When repetition is allowed, the problem can be treated as finding the number of ways to arrange 5 objects with 3 distinct types.

- This can be solved using the combination formula, where we choose 5 positions out of 7 (5 elements + 3 distinct types - 1).

- The binomial coefficient [tex]\binom{n}{r}[/tex] represents the number of ways to choose [tex]$r$[/tex] items from a set of [tex]$n$[/tex] items.

- In this case, we are choosing 5 positions (elements) from a set of 7 positions (5 elements + 3 distinct types - 1).

- The binomial coefficient can be expanded using factorials: [tex]\binom{n}{r} = \frac{n!}{r!(n-r)!}[/tex]

- Substituting [tex]n = 7$ and $r = 5[/tex], we get[tex]\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = 21[/tex]

Therefore, there are 21 different ways to select five elements in order from a set with three elements when repetition is allowed.