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3 years ago

Find y on this triangle

Mathematics

1 answer:

WARRIOR [948]3 years ago

5 0

Answer:

5√3 / 2

Step-by-step explanation:

tan 30 = ( 5 / 2 ) / y

1 / √3 = 5 / 2y

2y = 5√3

y = 5√3 / 2

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John haf 99 more stickers than Sally. Sally then gave John 36 more stickers. How many more stickers would John have than Sally i

densk [106]

Hi,

John haves 99 stickers. If Sally gave him 36 more that means john haves 99 + 36 = 135 more stickers than Sally.

Hope this helps.
r3t40

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4 years ago

Select the two binomials that are factors of this trinomial x2+10x+16

Zigmanuir [339]

So for this trinomial, I will be factoring by grouping. Firstly, what two terms have a product of 16x^2 and a sum of 10x? That would be 8x and 2x. Replace 10x with 2x + 8x:

$x^2+2x+8x+16$

Next, factor x^2 + 2x and 8x + 16 separately. Make sure that they have the same quantity inside the parentheses: $x(x+2)+8(x+2)$

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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,

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

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

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3 years ago

The sides of a whiteboard are 2 feet by 4 inches. How many 2 inch by 2 inch Post-it notes will be needed to cover the entire whi

weeeeeb [17]

Answer:

Step-by-step explanation:

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Blababa [14]

It's either a (the first one) or b (the second one), but if I had to choose I would choose b. If I'm wrong I'm so sorry I spent 10 minutes looking at this.

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4 years ago

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