Applied Algebra: Modeling and functions

Step 1

Let \(x\) stand for the size of the smallest angle. Then the medium-sized angle must be \(2x-15\text{.}\)

Step 2

Because the right angle is the largest angle, the sum of the smallest and medium-sized angles must be the remaining \(90\degree\text{.}\) Thus,

\begin{equation*} x + (2x - 15) = 90 \end{equation*}

Step 3

Solve the equation. Begin by simplifying the left side.

\begin{align*} 3x - 15 \amp = 90\amp\amp\blert{\text{Add 15 to both sides.}} \\ 3x \amp = 105\amp\amp\blert{\text{Divide both sides by 3.}}\\ x \amp = 35 \end{align*}

Step 4

The smallest angle is \(35\degree\text{,}\) and the medium-sized angle is \(2(35\degree) -15\degree\text{,}\) or \(55\degree\text{.}\)

Step 1

Let \(x\) represent the length of the shortest side, so that the third side has length \(2x\text{.}\)

Step 2

Substituting these expressions into the Pythagorean theorem, we find

\begin{equation*} x^2+(2x)^2=15^2 \end{equation*}

Step 3

This is a quadratic equation with no linear term, so we simplify and then isolate \(x^2\text{.}\)

\begin{align*} x^2+4x^2\amp =225\amp\amp\blert{\text{Combine like terms.}} \\ 5x^2\amp = 225\amp\amp\blert{\text{Divide both sides by 5.}} \\ x^2\amp = 45 \end{align*}

Taking square roots of both sides yields

\begin{equation*} x=\pm\sqrt{45}\approx\pm 6.708203932 \end{equation*}

Step 4

Because a length must be a positive number, the shortest side has length approximately \(6.71\) feet, and the third side has length \(2 (6.71)\text{,}\) or approximately \(13.42\) feet.

Let \(x\) represent the length of the third side of the triangle. By the triangle inequality, we must have that

\begin{equation*} x \lt 7 + 10, ~~~\text{ or }~~~ x \lt 17 \end{equation*}

Looking at another pair of sides, we must also have that

\begin{equation*} 10 \lt x + 7, ~~~\text{ or }~~~ x \gt 3 \end{equation*}

Thus the third side must be greater than \(3\) inches but less than \(17\) inches long.

Look at the diagram of the chalet at right. We can show that \(\Delta ABC\) is similar to \(\Delta ADE\text{.}\) Both triangles include \(\angle A\text{,}\) and because \(\overline{DE}\) is parallel to \(\overline{BC} \text{,}\) \(\angle ADE\) is equal to \(\angle ABC\text{.}\) Thus, the triangles have two pairs of equal angles and are therefore similar triangles.

1. Step 1.

   Let \(w\) stand for the width of the loft.

2. Step 2.

   First note that if \(FG = 8\text{,}\) then \(AF = 12\text{.}\) Because \(\Delta ABC\) is similar to \(\Delta ADE\text{,}\) the ratios of their corresponding sides (or corresponding altitudes) are equal. In particular,

   \begin{equation*} \frac{w}{24}=\frac{12}{20} \end{equation*}

3. Step 3.

   Solve the proportion for \(w\text{.}\) Begin by cross-multiplying.

   \begin{align*} 20w \amp = (12)(24)\amp\amp\blert{\text{Apply the fundamental principle.}}\\ w \amp = \frac{288}{20}= 14.4\amp\amp\blert{\text{Divide by 20.}} \end{align*}

4. Step 4.

   The floor of the loft will be \(14.4\) feet wide.

1. The formula for the volume of a right circular cylinder is \(V = \pi r^2h\text{.}\) If the height of the cylinder is \(6\) inches, then \(V = \pi r^2 (6)\text{,}\) or \(V = 6\pi r^2\text{.}\)

   

2. Substitute \(170\) for \(V\) and solve for \(r\text{.}\)

   \begin{align*} 170 \amp = 6\pi r^2\amp\amp \blert{\text{Divide both sides by }6\pi.}\\ r^2 \amp = \frac{170}{6\pi} \amp\amp\blert{\text{Take square roots.}}\\ r \amp = \sqrt{\frac{170}{6\pi}}\approx 3.00312 \end{align*}

   Thus, the radius of the can should be approximately \(3\) inches. A calculator keying sequence for the expression above is

   \begin{equation*} \boxed{\sqrt{~}} \,~\boxed{(} \, ~170 \, ~\boxed{\div} \, ~\boxed{(} \, ~6 \, ~\boxed{\pi} \, ~\boxed{)} \, ~\boxed{)} \, ~\boxed{\text{ENTER}} \end{equation*}

The box has six sides; we must find the area of each side and add them.

• The top and bottom of the box each have area \(lw\text{,}\) so together they contribute \(2lw\) to the surface area.
• The back and front of the box each have area \(lh\text{,}\) so they contribute \(2lh\) to the surface area.
• Finally, the left and right sides of the box each have area \(wh\text{,}\) so they add \(2wh\) to the surface area.

Thus, the total surface area is

Substitute \((2, -1)\) for \((x_1, y_1)\) and substitute \((4, 3)\) for \((x_2, y_2)\) in the distance formula to obtain

Substitute \((-2, 1)\) for \((x_1, y_1)\) and \((4, 3)\) for \((x_2, y_2)\) in the midpoint formula to obtain

The midpoint of the segment is the point \((\overline{x}, \overline{y}) = (1, 2)\text{.}\)

1. The graph of \((x - 2)^2 + (y + 3)^2 = 16\) is a circle with radius \(4\) and center at \((2, -3)\text{.}\) To sketch the graph, first locate the center of the circle. (The center is not part of the graph of the circle.)

   From the center, move a distance of \(4\) units (the radius of the circle) in each of four directions: up, down, left, and right. This locates four points that lie on the circle: \((2, 1)\text{,}\) \((2, -7)\text{,}\) \((-2, -3)\text{,}\) and \((6, -3)\text{.}\) Sketch the circle through these four points.

   

2. The graph of \(x^2 + (y - 4)^2 = 7\) is a circle with radius \(\sqrt{7} \) and center at \((0, 4)\text{.}\) From the center, move \(\sqrt{7} \text{,}\) or approximately \(2.6\text{,}\) units in each of the four coordinate directions to obtain the points \((0, 6.6)\text{,}\) \((0, 1.4)\text{,}\) \((-2.6, 4)\text{,}\) and \((2.6, 4)\text{.}\) Sketch the circle through these four points.

The center of the circle is the midpoint of its diameter. Use the midpoint formula to find the center:

Thus, the center is the point \((h, k) = (4, 2)\text{.}\) The radius is the distance from the center to either of the endpoints of the diameter, say the point \((7, 5)\text{.}\) Use the distance formula with the points \((7, 5)\) and \((4, 2)\) to find the radius.

Finally, substitute \(4\) for \(h\) and \(2\) for \(k\) (the coordinates of the center) and \(\sqrt{18} \) for \(4\) (the radius) into the standard form to obtain