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Katyanochek1 [597]

3 years ago

6

In a race, the second place finisher crossed the finish line 1 1/3 minutes after the winner. the third place finisher was 1 3/ m

inites behind the second place finisher. the thir place finisher took 34 2/3 minutes. How long did the winner tske

1 answer:

ss7ja [257]3 years ago

5 0

I think is answer is 9

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Guy ran at 13 km/hr. what was guys speed in MPH?

Assoli18 [71]

The guy ran at a speed of 8.07783 MPH

Here's how I got my answer:

1 km/hr. = 0.621371

So, if you do $0.621371 * 13$ you will get your answer: 8.07783 MPH

7 0

3 years ago

Read 2 more answers

GOOD REVIEW IF BOTH OF THE ANSWERS ARE CORRECT

Over [174]

Answer for 1:

I don't know the actual answer, but for Forms, you can press ctrl + U and find the answers in the code

Answer for 2:

x = 28

Step-by-step explanation:

Since the overall angle is 90°, and we already have 33, we subtract 33 from 90

90 - 33 = 57

So we have to fill in the other 57°

2(28) = 56

56 + 1 = 57

So the answer is x = 28

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

3 years ago

Helpplease ill give brainleist and points

Gre4nikov [31]

What is the question?

8 0

2 years ago

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Find the inequality represented by the graph

Kipish [7]

Answer:

3'4

Step-by-step explanation:

useibg rise over run

8 0

2 years ago

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