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2 years ago

Solve the equations: |x-1|/|x-1| = 1

Mathematics

2 answers:

Kryger [21]2 years ago

7 0

X = 0 ; 0-1 / 0-1 ; -1/-1= 1

Tanya [424]2 years ago

6 0

X = 2
The absolute value lines basically just serve as parentheses in this equation.
(2-1)/(2-1) = 1
(1)/(1) = 1

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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 $\rm \triangle OBA$ :

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

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

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3 years ago

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The 90th percentile corresponds to Z ≈ 1.28155.

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation

just olya [345]

Answer:

The particle will travel 6 feet in first 2 seconds.

Step-by-step explanation:

We have been given that a particle moves according to the velocity equation $v(t)= 6t^2-18t+12$ . We are asked to find the distance that the particle will travel in its first 2 seconds.

$s(t)=\int |v(t)|dt$

$s(t)=\int\limits^2_0 |6t^2-18t+12|dt$

Now, we will eliminate the absolute value sign as:

$s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt$

$s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1$

$s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1$

$s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))$

$s(2)=2-9+12-(0)-16+36-24-(-2+9-12)$

$s(2)=5-(0)-4-(-5)$

$s(2)=5-4+5$

$s(2)=6$

Therefore, the particle will travel 6 feet in first 2 seconds.

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2 years ago

For which equations below is x = –3 a possible solution? Check all that apply.

Fiesta28 [93]

|x|=3, |-x|=3, -|x|=-3. You can check each of the options one by one. First notice that the absolute value of some number can never be negative, so eliminate b, d and f. |-3|=3. |-(-3)|=|3|=3. -|3|=-3. So these three solutions are all correct answers.

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3 years ago