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3 years ago

I need help please!!

Mathematics

1 answer:

Montano1993 [528]3 years ago

3 0

Answer: D

Step-by-step explanation:

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Use the numbers 6 2 9 5 2 3 to make 36.

VashaNatasha [74]

Answer:

6+2+9+5+2+3+6+3 = 36

Step-by-step explanation:

If you like my answer than please mark me brainliest

5 0

3 years ago

If sin(2x) = cos(x + 30"), what is the value of x? 20° 30° 500 1100

ivanzaharov [21]

Answer:

2x=(x+30)

2x+x=30

3x=30

x=30÷3

x=10

7 0

3 years ago

Read 2 more answers

Two towers face each other separated by a distance = 15 m. As seen from the top of the first tower, the angle of depression of t

Crazy boy [7]

Answer:

Height of second tower = 17.32m

Step-by-step explanation:

I have attached a diagram depicting the question.

From the diagram, The first tower is depicted by side AEB and the second tower CD.

While d is the distance that separates the two towers and it's given as 15m.

Now, since the angle of depression of the second tower’s base is 60°, then for triangle BAC. Angle C = 60°.

Thus; using trigonometric ratios;

tan 60° = AB/AC.

This gives; AB = d*tan 60°

Similarly, for the triangle BED, BE = d*tan 30°

Since, AE = CD, thus ;

CD = AB − BE

CD = d (tan 60° − tan 30°)

CD = 15(1.7321 − 0.5774)

CD = 15 × 1.1547

CD ≈ 17.32 m.

So, height of second tower = 17.32 m

7 0

3 years ago

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

3 years ago

The cost of 17 m cloth is $77 5/7 find its cost per meter.

lukranit [14]

Answer:

Step-by-step explanation:

6 0

3 years ago