Currently, the area of Oak Park is represented by (9x2 + 6x + 9). The city will be expanding the park this summer. The area of the expansion is represented by 8x − 7. Which polynomial expression represents the total area of Oak Park after the expansion?

To know how much Marty pours into each bottle, divide the total juice amount by the number of bottles. His total of 2 4/5 quarts became 14/5 when turned into fifths for easier calculation. Dividing 14/5 by 2, we find that Marty pours 1 2/5 quarts into each bottle.

Marty has 2 4/5 quarts of juice. He is trying to equally distribute this juice into 2 bottles. To find out how much juice each bottle will contain, we divide the total amount of juice by the number of bottles.

Given that 1 full quart is made up of 5 fifths, 2 quarts are equivalent to 2x5=10 fifths. Marty has 2 full quarts and 4/5 of a quart, which amounts to 10+4=14 fifths of a quart. So, Marty has 14/5 quarts of juice in total.

If he pours the same amount into 2 bottles, then each bottle will contain half of the total amount. To find this out, we divide the total amount by 2, which gives us 14/5 ÷ 2 = 7/5 quarts. Therefore, Marty pours 7/5 quarts or 1 2/5 quarts of juice into each bottle.

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The solution to the system of equations 4x + 5y = -2 and 4x + 3y = 10 is x = 7 and y = -6. This is done by using the elimination method.

To solve this system of equations, you can use a method called elimination. The idea is to subtract the second equation from the first to eliminate one of the variables, allowing us to solve for the other.

Subtracting the second equation from the first gives you:

4x + 5y - (4x + 3y) = -2 - 10

This simplifies to: 2y = -12

So, y = -12 / 2 = -6

Then, substitute y = -6 into the first equation:

4x + 5(-6) = -2

This simplifies to 4x - 30 = -2, or 4x = 28

So, x = 28 / 4 = 7

So the solution to the system of equations is x =7, y = -6

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Given:

An open-top box with a square base has a surface area of 100 square inches.

Question:

The Process:

Part-1: The surface area

Let us arrange the equation to get the surface area of ​​the box with a square base. Recall that the box is without a lid and its surface area is 100 square inches.

[tex]\boxed{ \ Surface \ area = area \ of \ base + (4 \times area \ of \ rectangle) \ }[/tex]

[tex]\boxed{ \ x^2 + (4 \cdot x \cdot h) = 100 \ }[/tex]

[tex]\boxed{ \ x^2 + 4xh = 100 \ }[/tex]

From the above equation, we set it again so that "xh" is the subject on the left.

Both sides are subtracted by x².

[tex]\boxed{ \ 4xh = 100 - x^2 \ }[/tex]

Both sides are divided by 4.

[tex]\boxed{ \ xh = \frac{100 - x^2}{4} \ }[/tex] ... (Equation-1)

That is the strategy we have prepared.

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Part-2: The volume

[tex]\boxed{ \ Volume \ of \ the \ box = length \times width \times height \ }[/tex]

[tex]\boxed{ \ Volume = x \cdot x \cdot h \ }[/tex]

[tex]\boxed{ \ Volume = x^2h \ }[/tex] ...(Equation-2)

Substitution Equation-1 into Equation-2.

[tex]\boxed{ \ Volume = x(xh) \ }[/tex]

[tex]\boxed{ \ Volume = x \bigg( \frac{100 - x^2}{4} \bigg) \ }[/tex]

[tex]\boxed{ \ Volume = \frac{100x - x^3}{4} \ }[/tex]

[tex]\boxed{ \ Volume = 25x - \frac{1}{4}x^3 \ }[/tex]

Thus, an expression of the volume of the box as a function of the length of the edge of the base is [tex]\boxed{\boxed{ \ Volume = 25x - \frac{1}{4}x^3 \ }}[/tex]

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Part-3: The domain of volume

The value of volume must always be positive, i.e., V > 0.

[tex]\boxed{ \ 25x - \frac{1}{4}x^3 > 0 \ }[/tex]

Both sides are multiplied by 4.

[tex]\boxed{ \ 100x - x^3 > 0 \ }[/tex]

Both sides are multiplied by -1, notice the change in the sign of the inequality.

[tex]\boxed{ \ x^3 - 100x < 0 \ }[/tex]

[tex]\boxed{ \ x(x^2 - 100) < 0 \ }[/tex]

[tex]\boxed{ \ x(x - 10)(x + 10) < 0 \ }[/tex]

We get [tex]\boxed{ \ x = 0, \ x = 10, \ and \ x = - 10 \ }[/tex].

Consider the test of signs:

x(x - 10) (x + 10) is negative to the left of x = 10, and positive to the right of x = 10 on the number line.

Examples of tests:

Remember this form above, [tex]\boxed{\ x(x - 10)(x + 10) < 0 \ }[/tex], the value of the test result must be negative (because < 0).

Thus, the domain of the volume is [tex]\boxed{ \ 0 < x < 10 \ }[/tex] or [tex]\boxed{ \ (0, 10) \ }[/tex]

Keywords: an open-top box, with, a square base, has a surface area, 100 square inches, express, the volume, as a function, the length, edge, base, what, its domain, test of signs, substitution

1/3 * 84 = 28 were under 25

2/7 * 84 = 24 were at least 50

28 + 24 = 52

84-52 = 32 were between 25 and 50

Answer:

32 candidates are between 25 and 50.

Step-by-step explanation:

84 people applied for the job. [tex]\frac{1}{3}[/tex] of the candidates for a job were 25 years old or younger.

[tex]\frac{1}{3}[/tex] of 84 candidates = [tex]\frac{1}{3}[/tex] × 84 = 28 candidates

[tex]\frac{2}{7}[/tex] of the candidates were at least 50 years old.

[tex]\frac{2}{7}[/tex] of 84 candidates = [tex]\frac{2}{7}[/tex] × 84 = 24 candidates

age 25 and below + age 50 above = 28 + 24 = 52 candidates

Candidates between 25 and 50  =  84 - 52 = 32 candidates

32 candidates are between 25 and 50.

Answer: 6/11

Step-by-step explanation:

Open bracket

-69x+92+3x=56

Collect like term

-66x=-36

Divide both side by -66

x=6\11

It's C.!!! hope this helps :D!

We used De Morgan's laws to find the complement of the set formed by the intersection of the complement of A and B. We first found the complement of A, then the intersection of this set with B, and finally took the complement of this new set.

In set theory, De Morgan's laws relate the intersection and union of sets in a very elegant way. Here, (A' ∩ B)' represents the complement of the set formed by the intersection of the complement of A and B.

First, let's find A'. This is the set of all elements in U that are not in A. U = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } and A = { 0, 1, 4, 7, 9 } So, A' = { 2, 3, 5, 6, 8 }.

Now, let's find the intersection of A' and B, denoted by A' ∩ B. We have B = { 0, 4, 5, 6, 7 } so A' ∩ B = { 5, 6 }.

Finally, let's find the complement of this set, denoted by (A' ∩ B)'. This is the set of all elements in U that are not in A' ∩ B. So, (A' ∩ B)' = { 0, 1, 2, 3, 4, 7, 8, 9 }.

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Answer: False

Step-by-step explanation:

it’s just false I need points

Answer:

[tex]18a - 13.5[/tex]

Step-by-step explanation:

Jason's age is [tex]\frac{1}{2}[/tex] of his brother's age.

Brother age is [tex]12a-9[/tex]

Sum of their ages = Age of Jonson + age of his brother

Jonson's age = half of his brother age

[tex]Jonson's \ age=\frac{12a-9}{2}[/tex]

Sum of their ages =[tex]\frac{12a-9}{2}+12a-9[/tex]  

Take common denominator 2

[tex]\frac{12a-9}{2}+\frac{24a-18}{2}[/tex]  

[tex]\frac{36a-27}{2} [/tex]  

[tex]18a - 13.5[/tex]

It's accurate to say that 2×7 is not equal to 7×2, and the arrays provide a clear visual explanation of this mathematical concept.

Ms. Nenadal wrote 2×7 on the board, which is not equivalent to 7×2. The order of multiplication, as represented by the commutative property, states that changing the order of the factors does not change the product. In this case, 2×7 means adding 2 seven times, while7×2 means adding 7 two times.

Let's draw arrays to illustrate the difference:

For 2×7:

If we represent this as an array, it would have 2 rows and 7 columns, indicating 2 groups of 7.

Array for 2 × 7:

| * * * * * * * |

| * * * * * * * |

For 7×2:

On the other hand, 7×2 would be represented by an array with 7 rows and 2 columns, indicating 7 groups of 2.

Array for 7 × 2:

| * * |

| * * |

| * * |

| * * |

| * * |

| * * |

| * * |

While the product is the same (14) for both, the grouping and arrangement differ. Therefore, it's accurate to say that 2×7 is not equal to 7×2, and the arrays provide a clear visual explanation of this mathematical concept.

Answer:

x=6

Step-by-step explanation:

The large rectangle's dimensions are 3 times the dimensions of the small rectangle. If the small rectangle's dimensions are W for width and L for length, then the large rectangle's dimensions are 3W and 3L.

The subject of this question is Mathematics, specifically relating to geometry and ratios. You're asked to compare two rectangles: a large one and a small one. The large rectangle's dimensions (both the width and length) are three times the dimensions of the small rectangle. Let's assume the dimensions of the small rectangle are W (width) and L (length). Then, the dimensions of the large rectangle would be 3W and 3L, respectively. This is because the large rectangle is three times the size of the small rectangle in both width and length.

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The original price of the car was $5,500, which is derived from the mathematical equation based on the given information in the question.

The subject of this question is algebra, a branch of mathematics. The question gives us the selling price of a classic car and tells us that this price is $2500 less than three times its original price. We're asked to find the original price. Let’s denote the original price by x.

According to the problem, 3x (three times the original price) minus $2500 equals the selling price ($14,000). So, if we set this up as an equation, it looks like this: 3x - $2500 = $14,000.

To solve for x (the original price), we first need to add $2500 to both sides of our equation, giving us 3x = $16,500. Then, we divide each side by 3 to solve for x, arriving at x = $16,500 ÷ 3 = $5,500. Therefore, the original price of the car was $5,500.

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