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3 years ago

7

THIS IS DUE TODAY HELP PLEASE!!!

1 answer:

lozanna [386]3 years ago

5 0

This seems like a personal question

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Here are sketches of four triangles in each triangle the longest side is exactly 1cm the other length is given to 2 decimal plac

kakasveta [241]

ANSWER = 2.3

SJDEJNDJNJN

7 0

3 years ago

PLEASE HELP ME WITH THIS QUESTION, ILL BRAINLIEST YOU

PilotLPTM [1.2K]

Answer:

i think 105 is the correct answer Hope it helps you have a great day keep smiling be happy stay

8 0

3 years ago

You're flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 125 km south and 135 km east. Mt. Rainier is locat

Zanzabum

Answer:

5.12 minutes ≈ 5 minutes 7.2 seconds

Step-by-step explanation:

The flight path of the airplane is along the line ...

y = -125/135x = -25/27x

The perpendicular line through Mt. Rainier's location is ...

y = 27/25(x -56) -40

In standard form, these two equations are ...

25x +27y = 0
27x -25y = 2512

The point of closest approach has the coordinates that are the solution to these two equations:

(x, y) = (33912/677, 31400/677)

The distance d from the origin to this point is given by the Pythagorean theorem (distance formula) as ...

d = 1256√(2/677) . . . . . km

This distance can be converted to time using the speed of the airplane:

$(1256\sqrt{\dfrac{2}{677}}\,km)\times \dfrac{60\,min}{800\,km}=94.2\sqrt{\dfrac{2}{677}}\,min\approx 5.120019\,min$

5.12 minutes after departing JBLM you will be closest to Mt. Rainier.

_____

The attached graph shows the geometry of the problem.

3 0

4 years ago

A 15-ft ladder leans against a wall. The lower end of the ladder is being pulled away from the wall at the rate of 1.5 ft/sec. L

Aleks [24]

Answer:

The top of the ladder is sliding down at a rate of 2 feet per second.

Step-by-step explanation:

Refer the image for the diagram. Consider $\Delta ABC$ as right angle triangle. Values of length of one side and hypotenuse is given. Value of another side is not known. So applying Pythagoras theorem,

$\left ( AB \right )^{2}+\left ( BC \right )^{2}=\left ( AC \right )^{2}$

From the given data, $L=15\:ft=AC$ , $y=9\:ft=AB$ and $x=BC$

Substituting the values,

$\therefore \left ( 9 \right )^{2}+\left ( x \right )^{2}=\left ( 15 \right )^{2}$

$\therefore 81+x^{2}=225$

$\therefore x^{2}=225-81$

$\therefore x^{2}=144$

$\therefore \sqrt{x^{2}}=\sqrt{144}$

$\therefore x=\pm 12$

Since length can never be negative, so $x= 12$ .

Now to calculate $\dfrac{dy}{dt}$ again consider following equation,

$\left ( y \right )^{2}+\left ( x \right )^{2}=\left ( l \right )^{2}$

Differentiate both sides of the equation with respect to t,

$\dfrac{d}{dt}\left(y^2+x^2\right)=\dfrac{d}{dt}\left(l^2\right)$

Applying sum rule of derivative,

$\dfrac{d}{dt}\left(y^2\right)+\dfrac{d}{dt}\left(x^2\right)=\dfrac{d}{dt}\left(l^2\right)$

$\dfrac{d}{dt}\left(y^2\right)+\dfrac{d}{dt}\left(x^2\right)=\dfrac{d}{dt}\left(225\right)$

Applying power rule of derivative,

$2y\dfrac{dy}{dt}+2x\dfrac{dx}{dt}=0$

Simplifying,

$y\dfrac{dy}{dt}+x\dfrac{dx}{dt}=0$

Substituting the values,

$9\dfrac{dy}{dt}+12\times1.5=0$

$9\dfrac{dy}{dt}+18=0$

Subtracting both sides by 18,

$9\dfrac{dy}{dt}=-18$

Dividing both sides by 9,

$\dfrac{dy}{dt}= - 2$

Here, negative indicates that the ladder is sliding in downward direction.

$\therefore \dfrac{dy}{dt}= 2\:\dfrac{ft}{sec}$

7 0

3 years ago

Refer to the diagram below. Surveyors know that ∆PQR and ∆STR are similar. What is PQ, the distance across the lake?

Schach [20]

Bruh i dead had the same question lol. Sadly no one has answered yet.

7 0

4 years ago

Read 2 more answers