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3 years ago

What is 3 and 1 tenth as a desimal

Mathematics

1 answer:

Bess [88]3 years ago

5 0

Representing 3 and one tenth as decimal by 3.1

<u>Step-by-step explanation:</u>

Here 3 is the whole number.

1 tenth is nothing but $\frac{1}{10}$ .

$\frac{1}{10}$ = 0.1

When we write numerals with tenths using the decimals we can use a decimal point also the tenths positioned to the right of the decimal point. The tenth place is denoted as immediately right to the decimal point.

So it is represented as 3.1.

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soldi70 [24.7K]

Answer:

x=6. x=15

Step-by-step explanation:

13. x=6

16. x=15

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3 years ago

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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

3 years ago

The following table shows the distance from school as a function time

kykrilka [37]

Answer:

27,0 the distance away from the school

Step-by-step explanation:

Apply the formula for equation of a straight line;

y=mx+c where c is the y intercept , and m is the gradient

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Find the gradient by applying the formula;

$m=\frac{y_2-y_1}{x_2-x_1}$

Taking y₁=24, y₂=20, x₁=9, x₂=12

Then;

$m=\frac{20-24}{12-9} =\frac{-4}{3}$

write the equation of the function as ;

y=mx+c, c=36( the y-intercept when x=0) hence the equation is;

$y=mx+c\\\\y=\frac{-4}{3}x+36$

To get x-intercept, substitute value of y with 0

$y=mx+c\\\\\\y=-\frac{4}{3} +36\\\\\\0=-\frac{4}{3} x+36\\\\\frac{4}{3}x =36\\\\\\4x=36*3\\\\\\x=108/4=27$

This means that the x-intercept is 27 ,and this means you will require 27 minutes to cover the whole distance to the school.

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