A printer is printing photos. For every 12 photos, the printer takes 4 mins.

Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeee

natulia [17]

Answer:

77 m (nearest metre)

Step-by-step explanation:

Angle of Elevation

If a person stands at a point and looks up at an object, the angle between their horizontal line of sight and the object is called the angle of elevation.

To find how tall the flagpole is, model as a right triangle and solve using the tan trigonometric ratio.

Tan trigonometric ratio

$\sf \\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle

Given information:

Person height = 1.6 m
Horizontal distance (from person to flagpole) = 35 m
Angle of elevation = 65°

Therefore:

$\theta$ = 65°
O = height of flagpole less height of person
A = 35 m

As the person is 1.6 m tall, we model the line of sight as 1.6 m above ground level. Therefore, when calculating the height of the flagpole, we will need to add the height of the person to the value found using the trig ratio.

Substitute the given values into the formula and solve for O:

$\implies \sf \tan(65^{\circ})=\dfrac{O}{35}$

$\implies \sf O=35\tan(65^{\circ})$

Therefore the height of the flagpole is:

$\implies \sf 35\tan(65^{\circ})+1.6=76.65774222...$

$\implies \sf Height\:of\:flagpole=77\:m\:(nearest\:metre)$

Learn more about Angles of Elevation here:

brainly.com/question/27933160

8 0

2 years ago

You guys I need help with something like this again

Over [174]

In general an annual rate of $r$ compounded monthly will multiply the amount by a factor of $1+\frac{r}{12}$ each month.

Each month of 4.3% is a factor of $1+\frac{.043}{12}$ . That goes for five months, so is multiplied by itself five times, so gets a fifth power.

Each month of 13.7% is a factor of $1+\frac{.137}{12}$ . He pays this for 12-5=7 months, so it gets an exponent of seven.

We started the year owing $2600 of principal, so after the year we owe

$2600(1+\frac{.043}{12})^5(1+\frac{.137}{12})^7$ .

Choice A.

8 0

3 years ago

Someone help please please someone

aev [14]

(1-52/2+5)=(1-26+5)=(-25+5)=-20

continuing:

-20*2*6*6= -40*6*6=-240*6=-1440

result: -1440

6 0

3 years ago

Read 2 more answers

Calculate the area of the following figure correct to 2 decimal places if can show or explain work

grin007 [14]

The way we approach a question like this, is by looking at what portion of the circle we actually have to calculate the area of. We know that a circle is a shape that goes 360 degrees around a point. Therefore we know that, for example, a semicircle, or half a circle, would cover half the amount of degrees. This is important, because as well as covering half the number of degrees, it also covers half the area. For the figure in your picture, we are told that the portion missing from the circle is 90 degrees (shown by the right angle symbol in the center).
this means that 90 degrees worth of the circle is missing, so if we do 360÷90, we find that the answer equals $\frac{3}{4}$ . This means that we are dealing with $\frac{3}{4}$ of a circle. So, we simply find the area of the circle normally, and then multiply by $\frac{3}{4}$ :
The formula for area of a circle is $\pi r^{2}$ , so in this case, it can be said to be $\pi (10^{2} )$ , which is equal to 314.15
Now we simply take that area and multiply it by the fraction we established earlier:
314.15 x $\frac{3}{4}$ = 235.61
Therefore the area of the figure is 235.61 $cm^{2}$
I hope this helped and please remember to try and not only understand the answer, but the maths that worked it out too :))

3 0

4 years ago

A boat travel 312 miles each way downstream and back. The trip downstream took 13 hours. the trip back took 26 hours. What is th

zaharov [31]

Remember that $Speed= \frac{distance}{time}$ . Since the distance of the complete trip is 312 miles, the distance of each way is $\frac{312}{2} =156$ miles.
Let $S _{b}$ be the speed of the boat in still water and $S_{c}$ the speed of the current.

Now, for the downstream trip the boat is traveling with the current, so:
$S_{b} +S_{c} = \frac{156}{13}$
$S_{b} +S_{c} =12$ this will be our equation (1)

For the trip back the boat is traveling against the current, so:
$S _{b} -S_{c} = \frac{156}{26}$
$S_{b} -S_{c} =6$ this will be our equation (2)

Next, lets add equation (1) and equation (2) to get rid of $S _{c}$ :
$\left \{ {{S_{b}+S_{c} =12} \atop {+(S_{b}-S_{c} =6})} \right.$
$2S_{b} =18$
$S _{b} = \frac{18}{2}$
$S _{b} =9$

Finally, now that we know the speed of the boat in still water, lets replace that value in our equation (1) to find the speed of the current:
$9+S_{c} =12$
$S_{c} =12-9$
$S_{c} =3$

We can conclude that the speed of the boath in still water is 9 $\frac{mi}{h}$ , and the current of the stream is 3 $\frac{mi}{h}$ .

7 0

4 years ago