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2 years ago

Solve.

Mathematics

1 answer:

abruzzese [7]2 years ago

4 0

Answer:

$x=\frac{15}{4}=3\frac{3}{4}$

Equation:

$\frac{2}{3}x+\frac{5}{6} =\frac{10}{3}$

<h3>Step-by-step solution</h3>

Linear equations with one unknown

_____________________________________

1. Group all constants on the right side of the equation

$\frac{2}{3}\cdot x+\frac{5}{6}=\frac{10}{3}$

Subtract $5/6$ from both sides:

$\frac{2}{3}x+\frac{5}{6}-\frac{5}{6}=\frac{10}{3}-\frac{5}{6}$

Combine the fractions:

$\frac{2}{3}\cdot x+\frac{5-5}{6}=\frac{10}{3}-\frac{5}{6}$

Combine the numerators:

$\frac{2}{3}\cdot x+\frac{0}{6}=\frac{10}{3}-\frac{5}{6}$

Reduce the zero numerator:

$\frac{2}{3}\cdot x+0=\frac{10}{3}-\frac{5}{6}$

Simplify the arithmetic:

$\frac{2}{3}\cdot x=\frac{10}{3}-\frac{5}{6}$

Find the lowest common denominator:

$\frac{2}{3}\cdot x=\frac{10\cdot 2}{3\cdot 2}+\frac{-5}{6}$

Multiply the denominators:

$\frac{2}{3}\cdot x=\frac{10\cdot 2}{6}+\frac{-5}{6}$

Multiply the numerators:

$\frac{2}{3}\cdot x=\frac{20}{6}+\frac{-5}{6}$

Combine the fractions:

$\frac{2}{3}\cdot x=\frac{20-5}{6}$

Combine the numerators:

$\frac{2}{3}\cdot x=\frac{15}{6}$

Find the greatest common factor of the numerator and denominator:

$\frac{2}{3}\cdot x=\frac{5\cdot 3}{2\cdot 3}$

Factor out and cancel the greatest common factor:

$\frac{2}{3}\cdot x=\frac{5}{2}$

2. Isolate the x

$\frac{2}{3}\cdot x=\frac{5}{2}$

Multiply both sides by inverse fraction 3/2:

$\frac{2}{3}x\cdot \frac{3}{2}=\frac{5}{2}\cdot \frac{3}{2}$

Group like terms:

$\frac{2}{3}\cdot \frac{3}{2}x=\frac{5}{2}\cdot \frac{3}{2}$

Simplify the fraction:

$x=\frac{5}{2}\cdot \frac{3}{2}$

Multiply the fractions:

$x=\frac{5\cdot 3}{2\cdot 2}$

Simplify the arithmetic:

$x=\frac{15}{2\cdot 2}$

Simplify the arithmetic:

$x=\frac{15}{4}$

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Why learn this

Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends?

Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives.

__________________________

Terms and topics

Linear equations with one unknown

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3 years ago

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$\huge\text{Hey there!}$

$\large\textsf{-4 + 3x = 20}\\\rightarrow\large\textsf{3x - 4 = 20}\\\\\large\text{ADD 4 to BOTH SIDES}\\\large\textsf{3x - 4 + 4 = 20 + 4}\\\large\text{CANCEL out: \textsf{-4 + 4} because that gives you \textsf 0}\\\large\text{KEEP: \textsf{20 + 4}} \text{ because it gives you the \textsf{x-value}}}\\\\\large\textsf{20 + 4 = \bf 24}\\\\\large\text{DIVIDE 3 to BOTH SIDES}\\\\\mathsf{\dfrac{3x}{3}= \dfrac{24}{3}}\\\large\text{CANCEL out: }\mathsf{\dfrac{3}{3}}\large\text{ because that gives you \textsf 1}$

$\large\text{KEEP: }\mathsf{\dfrac{24}{3}}\large\text{ because it helps gives you the result of the \textsf{x-value}}\\\\\mathsf{x = \dfrac{24}{3}}\\\\\large\textsf{x= 24}\div \large\textsf{3}}\\\large\textsf{x = \bf 8}\\\\\boxed{\boxed{\huge\text{Therefore, your answer is: \textsf{x = 8}}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

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3 years ago

Read 2 more answers

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3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

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