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2 years ago

(a^2-ax+bx-ab)/(a^2+ax-bx-ab)*(a^2+ax+ab+bx)/(a^2-ax-ab+bx)

1 answer:

8 0

$\frac{a^2-ax+bx-ab}{a^2+ax-bx-ab} \cdot \frac{a^2+ax+ab+bx}{a^2-ax-ab+bx} = \frac{1}{a(a+x)-b(x+a)} \cdot (a(a+x)+b(a+x)) = \frac{1}{a-b} \cdot (a+b) = \frac{a+b}{a-b}.$

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Solve the following equation 5(y+4) =4(y+5)

kvv77 [185]

Answer:

Step-by-step explanation:

5(y+4) = 4(y+5)

5y+20 = 4y + 20

5y = 4y

y=0

6 0

3 years ago

Arianna is working two summer jobs, making $6 per hour walking dogs and making $10 per hour tutoring. In a given week, she can w

Mice21 [21]

Step-by-step explanation:

x = number hours dog walking

y = number hours tutoring

x + y <= 11

6x + 10y >= 80

the solution is the common area below the first line and above the second line.

the first line is above the second line (due to the y-intercept of 11 vs. 8) until the crossing point.

for the crossing point we use the regular equations :

from the first we get

x = 11 - y

using this in the second

6×(11 - y) + 10y = 80

66 - 6y + 10y = 80

4y = 14

y = 14/4 = 7/2 = 3.5

x = 11 - y = 11 - 3.5 = 7.5

so, she has to do at least 3.5 hours tutoring (and then 7.5 hours dog walking) for the minimum of $80 earning, all the way up to the maximum of 11 hours tutoring (and 0 dog walking) for $110 earning.

8 0

2 years 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 $\rm \triangle OBA$ :

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