A building is 139 ft tall.at noon it cast a shadow a 219- ft.what is the sun’s angle of elevation at that time.

Megan reads 2 pages of a book each morning

Makovka662 [10]

It’s A because 2+3=5 and 5x7=35

8 0

3 years ago

Helppppppp please hjbvkkkkkkkkk

Studentka2010 [4]

Do it by yourself ok

3 0

2 years ago

Read 2 more answers

Type the correct answer in each box. Use numerals instead of words. For this question, any answer that is not a whole number sho

egoroff_w [7]

Answer:

Step-by-step explanation:

Since we are not given the value of P, |Q|, R and S, we can as well assume values for them for the sake of this question.

Let P = 5, |Q| = 6, R =7 |S| = 2

Note that since Q and S are in modulus sign, they can return both positive and negative values.

P+Q = 5 + 6 (note that the positive value of Q is used since we need the greatest value of P+Q)

P+Q = 11

Hence the greatest value of P+Q is 11

For the least value of P+Q, we will use the negative value of Q as shown

P+Q = 5+(-6)

P+Q = 5-6

P+Q = -1

Hence the least value of P+Q is -1

Similarly:

R+S = 7 + 2 (note that the positive value of S is used since we need the greatest value of R+S)

R+S = 9

Hence the greatest value of R+S is 9

For the least value of R+S, we will use the negative value of S as shown

R+S = 5+(-2)

R+S = 5-3

R+S = 2

Hence the least value of P+Q is 2

NOTE THAT THIS ARE ASSUMED VALUES. ALL YOU NEED IS TO PLUG IN THE VALUES OF P, Q and R THAT YOU HAVE IN CASE THE VALUES DIFFERS.

5 0

3 years ago

Agent Bond is standing on a bridge, 13.5 m above the road below, and his pursuers are getting too close for comfort. He spots a

Scrat [10]

Answer:

Step-by-step explanation:

Eek! Let's give this a go. Things we know:

acceleration of Bond in free fall is -9.8 m/s/s

velocity of the truck is 25 m/s

displacement Bond will travel when he jumps is -10 m

What we are looking for is the time it will take him to hit the top of the truck, knowing that the truck can travel from one pole to the next in 1 second.

Our displacement equation is

Δx = v₀t + 1/2at²

Filling in we have

$-10=25t+\frac{1}{2}(-9.8)t^2$

Simplifying we get

$-10=25t-4.9t^2$

This is a quadratic that needs to be solved however you personally solve quadratics. When you do that, you find that the times it will take Bond to drop that displacement is either -.37 seconds or 5.47 seconds. Many things in physics can be negative, like velocity and acceleration, but time NEVER will be. So it takes Bond 5.5 seconds to drop to the roof of the moving truck. That means that he needs to jump when the truck is between the 5th and the 6th poles away from him.

Good luck with this!

Cheers!

6 0

3 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 $\rm \triangle OBA$ :

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

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 $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago