One way to solve this problem is to recognize that, in order for \(\frac{24}{30}\) to be greater than \(\frac{t}{24}\), t must be less than 24.

This is true because any fraction with a denominator larger than or equal to its numerator will always be less than a fraction whose numerator is larger than or equal to its denominator.

So, \(\frac{t}{24}\) is only less than \(\frac{24}{30}\) when t is less than 24.

Because you are asked for the largest integer value of t and you know that t must be less than 24, the correct answer must be 19.

You could also set the values equal to each other, cross-multiply and solve for t, as follows:

\(\frac{24}{30}=\frac{t}{24}\)

30t = 576

t = 19.2

The integer value that makes \(\frac{24}{30}=\frac{t}{24}\) greater than \(\frac{24}{30}=\frac{t}{24}\),therefore, is 19.

The equation of a circle is (x − h)² + (y − k)² = r², where h and k are the x and y-coordinates of the center of the circle and r is the radius.

Therefore, the equation of the circle centered at point (12, −5) with a radius of 4 is (x − 12)² + (y + 5)² = 16.

One way to solve this problem is to use SOH CAH TOA for angle A.

You are given that the sine of angle A is 3/5, which means that the ratio of the length of the side opposite angle A to the length of the hypotenuse is 3 to 5.

So, the ratio 3/5 is equal to the ratio BC/AC.

You are given that AC, the hypotenuse, is equal to 16, so now you can set up a proportion, cross-multiply,and solve for BC, as follows:

3/5 = BC/AC

5BC = 3AC

5BC = 3(16)

5BC = 48

BC = 9.6

To solve this problem, set the fixed charge to x and set the number of half miles to y.

Now, select two values from the table to create a set of equations, and solve for x and y, as follows:

x + 5y = 8.00

x + 6y = 8.50

The x values will cancel out, leaving y = 0.50,

which means that the cost per half-mile is $0.50; eliminate answer choice A.

Now, substitute.for y in one of the equations to solve for x:

x + 5y = 8

x + 5(.5) = 8

x = 8 − 2.5 = 5.50

One way to solve this problem is to recognize that ab² is a common factor of a²b² and ab³.

Therefore, it must also be a factor of 45, which can be factored as 5 × 3².

Now you can see that b could be 3, and since there can only be one correct answer, F must be it.

If you test a = 5 and b = 3, you see that the greatest common factor of 5² × 3² and 5 × 3³ is 5 × 3², which equals 45.

You are given certain ratios in this problem, so you can set up a proportion to find the length of the shortest side of the second triangle. The ratio of the longest sides of both triangles is \(\frac{15}{10}\).

Set the length of the shortest side of the second triangle to x so that the ratio of the shortest sides of both triangles is \(\frac{12}{x}\).

Now, set up a proportion, cross-multiply, and solve for x:

\(\frac{15}{10}=\frac{12}{x}\)

15x = 120

x = 8

You are given that x/y = 4, so x = 4y. Substitute 4y for x in the equation and solve, as follows:

x² − 12y²

= (4y)

2 − 12y²

= 16y² − 12y²

= 4y²

To solve this problem, first convert the information given into its mathematical equivalent, as follows (use b to represent the larger number):

The larger of two numbers exceeds twice the smaller number by nine: b = 2a + 9

The sum of twice the larger and five times the smaller number is 74: 2b + 5a = 74

Now, simply substitute the value of b into the second equation in order to solve for a:

2(2a + 9) + 5a = 74

To solve this problem, first recognize that, if n = m + 2, then m − n = −2.

Now, substitute this value into the second expression and solve:

(m − n)⁴ = (−2)⁴ = 16

To solve this problem, first expand the numerator and factor the denominator, then simplify, as follows:

\(\frac{(x − 13)(x − 13)}{(x − 13)(x + 13)}=\frac{(x − 13)}{(x + 13)}\)