Tom shoveled 12.5 yards of sand in 2.5 hours. At that rate how many yards did he shovel in one hour?

The variables as length are L and width is W and the system of equations is L = 2W - 3 and 42 = 2(L + W) and the length and width are 13 feet and 8 feet respectively.

A rectangle is a geometrical figure in which opposite sides are equal.

The angle between any two consecutive sides will be 90 degrees.

Area of rectangle = length × width.

As per the given total fencing = 42 feet

Let's say the length is L and the width is W.

As per the given,

L = 2W - 3

Perimeter = 2(length + width)

42 = 2(L + W)

21 = 2W - 3 + W

21 + 3 = 3W

W = 8 feet

And, L = 2(8) - 3 = 13 feet

Hence "The variables as lengths are L and width is W and the system of equations is L = 2W - 3 and 42 = 2(L + W) and the length and width are 13 feet and 8 feet respectively".

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Answer: B.)3.34

I did it on usatestprep

Step-by-step explanation:

Adam bought 26.14 ounces of orange juice.

To solve the problem, we need to determine how many ounces of orange juice Adam bought with $3.42, given that the cost per ounce is $0.12. We can set up a simple proportion to find the answer.

First, we convert the cost of orange juice per ounce from a fraction to a decimal. The cost is $0.12 per ounce, which can also be written as [tex]$\frac{12}{100}$[/tex] dollars per ounce. Simplifying this fraction gives us [tex]$\frac{3}{25}$[/tex] dollars per ounce.

Next, we convert the total amount spent by Adam, $3.42, into a fraction. This is equivalent to [tex]$\frac{342}{100}$[/tex] dollars.

Now, we set up the equation to find the number of ounces (let's call it [tex]$x$[/tex]) that Adam bought:

[tex]\[ \frac{3}{25} \times x = \frac{342}{100} \][/tex]

To solve for [tex]$x$[/tex], we multiply both sides of the equation by the reciprocal of [tex]$\frac{3}{25}$[/tex], which is [tex]$\frac{25}{3}$[/tex]:

[tex]\[ x = \frac{342}{100} \times \frac{25}{3} \][/tex]

Multiplying the numerators and denominators separately, we get:

[tex]\[ x = \frac{342 \times 25}{100 \times 3} \][/tex]

Simplifying the right side of the equation by canceling out common factors (25 cancels with 100 to become 4, and 3 cancels with 342 to become 114), we have:

[tex]\[ x = \frac{114 \times 4}{4} \][/tex]

This simplifies to:

[tex]\[ x = 114 \][/tex]

However, we must remember that the original amount spent was $3.42, not $342. Therefore, we must divide our result by 100 to account for the cent conversion:

[tex]\[ x = \frac{114}{100} \][/tex]

[tex]\[ x = 1.14 \][/tex]

This result represents the number of ounces Adam bought in addition to the whole number of ounces he could buy with whole dollars. Since $3.00 would buy [tex]$\frac{3}{0.12}$[/tex] ounces, we need to add this to our 1.14 ounces to get the total:

[tex]\[ x_{total} = 1.14 + \frac{3}{\frac{12}{100}} \][/tex]

[tex]\[ x_{total} = 1.14 + \frac{3 \times 100}{12} \][/tex]

[tex]\[ x_{total} = 1.14 + \frac{300}{12} \][/tex]

[tex]\[ x_{total} = 1.14 + 25 \][/tex]

[tex]\[ x_{total} = 26.14 \][/tex]

The question relates to house appreciation and quality in mathematics. The house's value appreciates each year, an occurrence that Heather and Joel could monitor. House appreciation allows them to determine the annual interest rate and the value of their house after a particular period, and Ben and Freda's specifics help illustrate this situation.

The question pertains to the concept of housing appreciation and equity in Mathematics. In this situation, Heather and Joel will observe that the value of their home increases annually due to appreciation. The value of their home after a certain period can be modeled using an exponential equation, taking the initial amount as the purchasing price, and the rate as the annual appreciation percentage.

For example, looking at Freda's situation, she bought a house for $150,000. Now, if she were to sell it, she would get $250,000. This increase in price from $150,000 to $250,000 would demonstrate the appreciation of her house over time. Similarly, Ben bought a house for $100,000 and borrowed the rest after a 20% down payment. As he pays the loan, his equity in the house increases and the house's increase in value also appreciates his equity.

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The total surface area of a cone can be found using the formula A = πr(r + l), where A is the total surface area, r is the radius of the base, and l is the slant height. In this case, the correct answer is 115π cm².

The total surface area of a cone can be found using the formula A = πr(r + l), where A is the total surface area, r is the radius of the base, and l is the slant height.

In this case, the correct answer is 115π cm². Let's calculate it:

Therefore, the correct answer is 115π cm².

To determine the function which has the largest value at x=3,

We will calculate the value of each function by substituting the value of x=3 in each of the given function.

Let us consider the first function,

[tex] c(x)=3x^{2}+5x+22 [/tex]

[tex] c(3)=3\times 3^{2}+5 \times 3+22 [/tex]

[tex] c(3)=27+15+22 [/tex]

[tex] c(3)=64 [/tex]

Let us consider the second function,

[tex] j(x)=12x [/tex]

[tex] j(x)=12 \times 3 =36 [/tex]

Let us consider the third function,

[tex] a(x)=9x [/tex]

[tex] a(x)=9 \times 3=27 [/tex]

Therefore, the function c(x) has the largest value at x=3.

The answer is 3

To determine which function has the largest value when x = 3, we need to substitute x with 3 into each function and calculate their values.

c(x) = 3x2 + 5x + 22 => c(3) = 3(3)2 + 5(3) + 22 = 27 + 15 + 22 = 64

j(x) = 12x => j(3) = 12(3) = 36

a(x) = 9x => a(3) = 9(3) = 27

The function c(x) has the largest value of 64 when x = 3.

Answer: I hope this is helpful

Step-by-step explanation:

10LCM of 5 and 10 is 10. LCM, also known as Least Common multiple or Lowest common multiple, is the smallest or the least positive integer that is divisible by the given set of numbers. In the given set of numbers 5 and 10, 10 is the least number that is divisible by both 5 and 10.

To solve the equation 3x = 42, divide both sides by 3, which results in x = 14. This means you would pay $14. The process is similar to isolating a variable in quadratic equations, where different methods may be employed.

Given the equation 3x = 42, we need to isolate the variable x to find its value. This is accomplished by dividing both sides of the equation by 3, which is the coefficient of x. Therefore, the equation becomes x = 42 / 3.

By performing this simple division, we find that x equals 14. So, if x represents the amount you must pay, then you would pay $14.

The process of isolating the variable is essential for solving not only simple equations but also more complex ones, such as quadratic equations which take the form ax² + bx + c = 0.

When dealing with quadratic equations, you can employ the quadratic formula or other methods like factoring or completing the square, depending on the nature of the coefficients and the terms present in the equation.

The amount of salt that is in the tank after 20 minutes is;

A(20) = 267.56 Oz

Initial amount of salt in tank; A(0) = 40 Ounces

A solution containing 4 oz salt per gallon is pumped into the tank at a rate of 5 gal/min.

This means rate in oz/min = 4/1 × 5/1 = 20 oz/min

A'(t) = 20 - [(A(t)/(100)) × 5/1]

A'(t) = 20 - A(t)/20

Rearranging gives;

A'(t) + A(t)/20 = 20

Multiplying through by the integrating factor gives;

A'(t) [tex]e^{t/20}[/tex] + (A(t)/20) [tex]e^{t/20}[/tex] = 20[tex]e^{t/20}[/tex]

A(t) [tex]e^{t/20}[/tex] = 400[tex]e^{t/20}[/tex]  + C

Divide through by [tex]e^{t/20}[/tex] to get;

A(t) = 400 + C [tex]e^{-t/20}[/tex]

At initial condition of A(0) = 40, we have;

40 = 400 + C

C = -360

Thus; at time (t), the amount of salt left is given by;

A(t) = 400 - 360 [tex]e^{-t/20}[/tex]

A(20) = 400 - 360 [tex]e^{-20/20}[/tex]

A(20) = 400 - 132.44

A(20) = 267.56 Oz

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