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3 years ago

2b^2+4b–2=0 what is the answer?

1 answer:

s2008m [1.1K]3 years ago

5 0

$2b^2+4b-2=0\\
2(b^2+2b-1)=0\\
2(b^2+2b+1-2)=0\\
2((b+1)^2-2)=0\\
2(b+1)^2-4=0\\\
2(b+1)^2=4\\\
(b+1)^2=2\\
b+1=-\sqrt2 \vee b+1=\sqrt2\\
b=-1-\sqrt2 \vee b=-1+\sqrt2

$

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Simplify the expression. Write the answer using scientific notation.

bonufazy [111]

Answer:

C. 2.5×10^-27

Step-by-step explanation:

(2×10^-13)^-2 = 2^-2 × 10^(13×(-2)) = 1/4 × 10^-26

= 0.25 × 10^-26 = 2.5 × 10^-27

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3 years ago

You roll a fair six-sided die. The die shows an odd number or a number greater than two.

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Step-by-step explanation:

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3 years ago

Please help, see attachment for equation!

Alekssandra [29.7K]

Answer:

The answer to your question is the last option

Step-by-step explanation:

Quadratic equation

2 = - x + x² - 4

Order the equation from the highest power to the lowest power. Do not consider 2 because it is not consider in the options given.

x² - x - 4 = 0

Identify a, b and c

(1) x² -(1) x - 4 = 0

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Substitution

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3 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!

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3 years ago

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Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers