Solve y=3x-4 and y=2/5x 9

The value of\[\left( 1-{1\over3} \right)\left( 1-{1\over4} \right)\left( 1-{1\over5} \right)....\left( 1-{1\over n} \right)\]is

Oliga [24]

Example 1

verbose explicit high3 plus 4 cross 2 minus minus 2 equals 3 plus 8 plus 2 equals 1 3verbose explicit high semantics3 plus 4 times 2 minus negative 2 equals 3 plus 8 plus 2 equals 13verbose explicit high semantics high3 plus 4 times 2 minus negative 2 equals 3 plus 8 plus 2 equals 13

For most fractions, the beginning is indicated with "start fraction", the horizontal line is indicated with "over", and the end of the fraction is indicated by "end fraction". For the semantic interpretation, most numeric fractions are spoken as they are in natural speech. Also if a number is followed by a numeric fraction, the word "and" is spoken in between.

6 0

2 years ago

7 to the 6th power = n to the 8th power what is n

Valentin [98]

Answer:

n=7 3/4, -7 3/4

n=4.30351707...,-4.30351707...

8 0

3 years ago

2y = 3x + 4 2x = -3y - 7

anygoal [31]

Answer:

x= -2.........&.........y= -1

Step-by-step explanation:

(x,y)=(-2,-1)

7 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 $\rm \triangle OBA$ :

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