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2 years ago

Please help with this question!

Mathematics

1 answer:

Naddika [18.5K]2 years ago

6 0

I am pretty sure the answer is 14

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%20%5E%208%7D%7B5%20%5E%203%7D" id="TexFormula1" title="\frac{5 ^ 8}{5 ^ 3}" alt="\

Vlada [557]

Step-by-step explanation:

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7 0

2 years ago

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Solve x 3(x+1)=-2(x-1)-4

Nataly [62]

Answer:

x= -1/3 or -2

Step-by-step explanation:

3x^2 +3x= -2x +2-4

3x^2 + 5x +2=0

by factorization method

3x(x+2) 1(x+2)=0

(3x+1)(x+2)

3 0

3 years ago

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PLEASE HELP<br><br> Log,3 7=x is equivalent to which of the following (choose all correct answers)

ELEN [110]

Answer:

should be 3^X = 7 and X = log7/log3

Step-by-step explanation:

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2 years ago

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There are 10 decimeters in every meter. If a piece of fabric is 32 decimeters long, how long is it when measured in meters?

lara31 [8.8K]

10 dm = 1 m
32 dm = 32/10 = 3.2 m
Answer: 3.2 m

8 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

2 years ago