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2 years ago

Find one value of xxx that is a solution to the equation: (5x+2)^2+15x+6=0(5x+2) 2 +15x+6=0

1 answer:

5 0

An equation is formed of two equal expressions. The solution of the equation (5x+2)² + 15x + 6=0 is -0.4 and -1.

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

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The solution to the equation,

(5x+2)² + 15x + 6=0

25x² + 4 + 20x + 15x + 6 = 0

25x² + 35x + 10 = 0

Plotting the equation on the graph, the value of x is,

x = -0.4, -1

Hence, the solution of the equation (5x+2)² + 15x + 6=0 is -0.4 and -1.

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Simplify the expression 8xy + 6.3h + 7xy - xy - 2h

solong [7]

Answer:

14xy+4.3h

Step-by-step:

= 8xy + 6.3h + 7xy - xy + -2h

= (8xy + 7xy + -xy) + (6.3h + -2h)

=14xy + 4.3h

6 0

3 years ago

Please find the result !

Sliva [168]

Answer:

$\displaystyle - \frac{1}{2}$

Step-by-step explanation:

we would like to compute the following limit:

$\displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right)$

if we substitute 0 directly we would end up with:

$\displaystyle\frac{1}{0} - \frac{1}{0}$

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

now notice that after simplifying we ended up with a<em> </em><em>rational</em><em> </em>expression in that case to compute the limit we can consider using L'hopital rule which states that

$\rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right)$

thus apply L'hopital rule which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right)$

simplify which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right)$

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

$\displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } }$

simplify which yields:

$\displaystyle - \frac{1}{2}$

finally, we are done!

6 0

3 years ago

Read 2 more answers

Pls help im in dire need cause bainly doesnt got this yet

ohaa [14]

$\sf -3\le x\le3$ as the line has the ends which has closed dot to indicate that the endpoint is part of the solution. Also its between the points -3 and 3.