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3 years ago

Select the correct answer. What is the product of and ? A. 9/5 B. 2 C.175/30 D/29

Mathematics

2 answers:

Sergio039 [100]3 years ago

6 0

What is the product of -3 1/3 and -8 7/10

C.175/30

jonny [76]3 years ago

4 0

Step-by-step explanation:

the answer is c i took the test btw

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PLEASE HELP. Prove that the value of the expression: b (36^5−6^9)(38^9−38^8) is divisible by 30 and 37.

pishuonlain [190]

Answer:

Step-by-step explanation:

First factor the second binomial

38^9 - 38^8 = 38^9 (38 - 1 ) = 38^9(37)

So there's your 37. This whole expression is divisible by 37

Now do the first binomial

36^5 - (36^4)*6

36^4(36 - 6)

36^4(30) and there's your thirty.

That first term is going to cause a bit of trouble showing that 36^4 * 6 = 6^9

6(36^4) = 6(6^2)^4 = 6^1 * 6^8 = 6^9

So this factors into 30*(36^4)*37*38^9

3 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

2 years ago

What is the true solution to In 20+ In 5= 2 In x

amid [387]

X would equal 10. Hope this helps!

7 0

3 years ago

Give a rule to the function. {(2,5), (3,10), (4,17), (√3, 4)}

Luda [366]

$f(x)=x^2+1$

3 0

3 years ago

A Discount clothing store is selling a $72 sport jacket for $24 off the regular price. What is the discount rate?

mr Goodwill [35]

72-24=48

x% of 72=48

(/72)x/100 *72=48 (/72)

(*100 ) x/100=2/3 (*100)

x= 66.67%

100% -66.67%=33.33% discount

7 0

3 years ago