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3 years ago

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A magician performing a magic act has asked for 2 volunteers from the audience to help with his routine. If there are 250 people

in the audience, which of the following expressions represents the number of possible pairs of volunteers?

2 answers:

Kay [80]3 years ago

6 0

Solution:

Number of people in the audience = 250

Number of Volunteers to be chosen from 250 volunteers = 2

As this can be done by choosing 2 people out of 250 peoples , the order is not important.

So , we will use combinatorics for this method, because order does not matter here.

So Possible combination= $_{2}^{250}\textrm{C}$

= $\frac{250!}{248!\times 2!}=249 \times 125$

= 31125 ways

RUDIKE [14]3 years ago

5 0

The answer is 125. If you were to pair 250 people in 2, that would be 250 divided by 2. And 250 divided by 2 is 125.

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stepan [7]

Answer:

The answer is 21 units²

Step-by-step explanation:

To find the area of a right triangle you use the formula A=ab/2

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2 years ago

The cooking class used 18 cups of flour to make 8 identical loaves of bread. How many cups of flour were needed for each loaf ?

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2 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!

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3 years ago

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morpeh [17]

Answer:

see the explanation

Step-by-step explanation:

we know that

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substitute

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Part 2) we have

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Dima020 [189]

The matching is given below.

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<h3>What is an exponent?</h3>

Let a be the initial value and x be the power of the exponent function and b be the increasing factor. The exponent is given as

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Then the matching is given below.

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More about the exponent link is given below.

brainly.com/question/5497425

#SPJ1

5 0

1 year ago