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Anni [7]
3 years ago
7

1.) A guy wire (a wire used to stabilize something) 115 feet long runs from

Mathematics
1 answer:
liberstina [14]3 years ago
8 0

Answer:

62.2°

Step-by-step explanation:

If a wire used to stabilize something is 115 feet long runs from

the top of a radio tower to a point on the ground 50 feet from the tower's base, then;

The set up will form a right angle triangle as shown

The Adjacent side = The base of the tower = 50 feet

The hypotenuse = length of the wire = 115feet

The Height of the tower = Opposite side of the set up.

theta = angle that the wire make with the ground

Using the SOH CAH TOA trigonometry Identity:

According to CAH;

Cos(theta) = Adjacent/Hypotenuse

Cos(theta) = 50/115

Cos theta = 0.435

theta = arccos 0.435

theta = 62.2°

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Answer:

A. No real solution

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Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

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Refer to the attachment for the graph as well .

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