Solve the two equations, 3x+ 2y= –2 and 5x– y= 14, as a system of equations.

To do this, multiply the second equation by 2 and then add it to the first equation:

\(\frac{\begin{matrix} & 3x & + & 2y & = & -2\\ + & 10x & - & 2y & = & 28 \end{matrix}}{\begin{matrix} & 13x & \;\; & & \;\;\; = & 26 \end{matrix}}\)

Now solve for x:

x = 2

Replace x with 2 in either equation and solve for y:

3(2) + 2y = –2

6 + 2y = –2

2y = –8

y = –4

At the point of intersection, x = 2 and y = –4, so this point is (2, –4).

The box has dimensions of 3 and 9 and a volume of 135, so plug these values into the formula for the volume of a box:

V = lwh

135 = (3)(9)h

135 = 27h

5 = h

So the remaining dimension of the box is 5.

The two longest dimensions are 5 and 9, so the area of the largest side is 5 × 9 = 45.

Factor the left side of the equation:

t² – 6t + 8 = 0

(t – 2)(t – 4) = 0

Split this equation into two equations and solve:

t – 2 = 0 or t – 4 = 0

t = 2 ; t = 4

Thus, a = 4 and b = 2.

So a – b = 4 – 2 = 2.

∠p and ∠r are opposite angles on the parallel lines, so they total 180°, which rules out Choice (A).

The same is true of ∠q and ∠s,ruling out Choice (C).

∠t and ∠u are supplementary, so they add up to 180°, which rules out Choice (E).

Finally, ∠r, ∠s, and ∠t are all vertical angles with the three interior angles of a triangle, so they total 180°, ruling out Choice (D).

By elimination, the correct answer is Choice (B).

Let the two integers equal x and y,and then create the following equation and simplify:

xy + 4 = 40

xy = 36

So x and y are a pair of integers that equal 36.

Try adding all possible combinations of two integers that multiply out to 36:

1 × 36 = 36; 1 + 36 = 37
2 × 18 = 36; 2 + 18 = 20
3 × 12 = 36; 3 + 12 = 15
4 × 9 = 36; 4 + 9 = 13
6 × 6 = 36; 6 + 6 = 12

This list of sums includes 12, 13, 15, and 20, but not 18.

Thus, no pair of integers both satisfies the original equation and adds up to 18.

The problem states that f(x) = 4x² – 5x – 5, so

f(–5) = 4(–5)² – 5(–5) – 5 = 100 + 25 – 5 = 120

You also know that g(x) = 2^x – 4, so

g(6) = 2⁶ – 4 = 64 – 4 = 60

Therefore: \(\frac{g\left ( 6 \right )}{f\left ( -5 \right )} = \frac{60}{120} = \frac{1}{2}\)

Solve the equation pq – 3r = 2 in terms of p and r by isolating the q on one side of the equal sign as follows:

pq - 3r = 2

pq = 2 + 3r

\(q = \frac{3r + 2}{p}\)

The secant of an angle equals the hypotenuse over the adjacent angle, so

sec n = \(\frac{H}{A} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2}\)

To begin, find the area of each of the two sections by multiplying the lengths of the sides:

30 feet ×40 feet = 1,200 square feet

60 feet ×80 feet = 4,800 square feet

The total area that needs sod has an area of 1,200 + 4,800 = 6,000 square feet.

Each individual square is 2 feet by 2 feet, so each has an area of 4 square feet.

Because all the numbers are even, there will be no waste when the squares of sod are placed.

Therefore, you just have to find the number of squares needed by dividing:

6,000 ÷ 4 = 1,500

Plug the values for the coordinates into the formula for the slope of a line:

Slope = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-3 -3 }{4 - 1} = \frac{-6}{3} = -2\)