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Aloiza [94]
3 years ago
10

In triangle ABC, AD is a median. FE is a straight line parallel to BC cutting the remaining sides AB at F and AC at E and cuttin

g the median at X.
If ED bisects angle ADC:


Triangle AFX is similar to Triangle ABD and that Triangle AEX is similar to Triangle ACD.


Using above information prove that angle FDE=90
Mathematics
1 answer:
daser333 [38]3 years ago
4 0

Answer:

Step-by-step explanation:

since AD is a median it implies that triangle ABC is bisected to two equal right angled triangle which are ADB and ADC.

FE is parrallel to BC and cuts AB at F and AC at E shows that there are two similar triangles formed which are AFE and ABC.

Recall that ADC is a right angled triangle, ED bisects a right angled triangle the the ADE = 45^{o}.

Now, Let FD bisect angle ADB,

  then ADF = 45^{o} too.

Since AFX is similar to Triangle ABD and that Triangle AEX is similar to Triangle ACD, then EDX is similar to FDX

FDE = ADF + ADE = 90^{o}

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The Ferris wheel is the most popular ride. In 1 hour, the Ferris wheel had 240 riders. If that was 60% of the total number of ri
Vesnalui [34]

Answer:

400 riders

Step-by-step explanation:

The Ferris wheel had 400 riders.

400*0.60=240.

4 0
3 years ago
Read 2 more answers
A tortoise and a hare are competing in a 2000-meter race. The arrogant hare decides to let the tortoise have a 550-meter head st
andrew-mc [135]

Answer:

a) distance covered by hare d1 = 8t

b) distance covered by tortoise d2 = 5t + 550

c) ∆d = 550 - 3t

Step-by-step explanation:

Given;

Speed of hare u = 8m/s

Speed of tortoise v = 5 m/s

Initial distance of tortoise d0 = 550 m

a) using the equation of motion;

distance covered = speed × time + initial distance

d = vt + d0

For hare;

d0 = 0

Substituting the values;

d1 = 8t + 0

d1 = 8t

b)using the equation of motion;

distance covered = speed × time + initial distance

d2 = vt + d0

For tortoise;

d0 = 550m

Substituting the values;

d2 = 5t + 550

d2 = 5t + 550 m

c) the number of meters the tortoise is ahead of the hare.

∆d = distance covered by tortoise - distance covered by hare

∆d = d2 - d1

Substituting the values;

∆d = (5t + 550) - 8t

∆d = 550 - 3t

6 0
2 years ago
Helpppppp plsssssssss
bekas [8.4K]

Answer:

C

Step-by-step explanation:

The right triangle on the right side of the figure has a height of 6 (two same sides lengths) and a base of 3.

x is the hypotenuse (side opposite of 90 degree angle).

We can use the pythagorean theorem to find x. The pythagorean theorem tells us to square each leg (height and base) of the triangle and add it. It should be equal to the hypotenuse square.

For this triangle it means, we square 6 and 3 and add it. It should be equal to x squared. Then we can solve. Shown below:

6^2 + 3^2 = x^2\\36 + 9 = x^2\\45 = x^2\\x=\sqrt{45}

<em>Now we can use property of radical  \sqrt{x}\sqrt{y}  =\sqrt{x*y}  to simplify:</em>

x=\sqrt{45} \\x=\sqrt{5*9} \\x=\sqrt{5} \sqrt{9} \\x=3\sqrt{5}

Correct answer is C

6 0
3 years ago
Read 2 more answers
Help me with differentation and integration please!!
Marina86 [1]

Answer:

See below

Step-by-step explanation:

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

Recall

\dfrac{d}{dx}\tan x=\sec^2

Using the chain rule

\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}

such that u = \tan x

we can get a general formulation for

y = \tan^n x

Considering the power rule

\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}

we have

\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x

therefore,

\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x

Now, once

\sec^2 x - 1= \tan^2x

we have

3\tan^2x \sec^2x =  3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x

Hence, we showed

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

================================================

For the integration,

$\int \sec^4 x\, dx $

considering the previous part, we will use the identity

\boxed{\sec^2 x - 1= \tan^2x}

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx $

Considering u = \tan x

and then du=\sec^2x\ dx

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx  = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0
2 years ago
CAN ANYONE PLEASE HELP ME !!
Ivan

Answer:

The answer is 9.

Step-by-step explanation:

8 0
2 years ago
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