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3 years ago

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

Mathematics

1 answer:

77julia77 [94]3 years ago

6 0

The square root is 8 the comment above is how you do the work^

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How to make 27/102 into a simplest form and 16/32 into simplest form?

Reil [10]

My caculator got 9/34

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3 years ago

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Please read the attachment below :)

SVEN [57.7K]

Answer:

y=7x

Step-by-step explanation:

y-int=0

m=35/5=m=7

remember.... rise over run - 35/5 is equal to 7,so the slope is 7

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3 years ago

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X = ? y = ? 16 45 degrees

mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}$

For x, we will be using the Cosine Function:

$\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}$ $(16)Cosine(45^{\circ})\text{ = x}$ $(16)(\frac{1}{\sqrt[]{2}})\text{ = x}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = x}$

Therefore, x = 8√2.

For y, we will be using the Sine Function.

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}$ $\text{ (16)Sine }(45^{\circ})\text{ = y}$ $\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = y}$

Therefore, y = 8√2.

5 0

1 year ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .