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2 years ago

What expression below is equivalent to -71n -

Mathematics

2 answers:

noname [10]2 years ago

7 0

<h3>Answer: Choice A) -29 - 71n</h3>

Reason:

An expression in the form -a-b is the same as -a + (-b)

Recall we can add two numbers in any order we want. For instance, 2+3 = 3+2 since both sides are equal to 5. This is the commutative property of addition.

Therefore, the -a + (-b) is the same as -b + (-a) which in turn becomes -b - a

In short: -a - b = -b - a

This rule lets us say that -71n - 29 = -29 - 71n

sergiy2304 [10]2 years ago

6 0

Answer:

Step-by-step explanation:

both numbers are negative, you can rearrange them in any order provided you keep the sign

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3 years ago

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3 years ago

Read 2 more answers

How does one integrate the following:

grin007 [14]

Consider substituting $u=4+xy$ , so that $\mathrm du=x\,\mathrm dy$ . Then

$\displaystyle\int_{x=a}^{x=b}\int_{y=a}^{y=b}\frac x{(4+xy)^2}\,\mathrm dy\,\mathrm dx=\int_{x=a}^{x=b}\int_{u=4+ax}^{u=4+bx}\frac1{u^2}\,\mathrm du\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(-\frac1u\right)\bigg|_{u=4+ax}^{u=4+bx}\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(\frac1{4+ax}-\frac1{4+bx}\right)\,\mathrm dx$

Then by similar substitutions, you can easily find that you end up with

$\dfrac1a\ln|4+ax|-\dfrac1b\ln|4+bx|\bigg|_{x=a}^{x=b}$
$=\dfrac1a\ln|4+ab|-\dfrac1b\ln|4+b^2|-\dfrac1a\ln|4+a^2|+\dfrac1b\ln|4+ab|$
$=\dfrac1a\ln\left|\dfrac{4+ab}{4+a^2}\right|+\dfrac1b\ln\left|\dfrac{4+ab}{4+b^2}\right|$

Of course, this all assumes that the integrand is continuous over the domain of integration, which would require that $a,b$ are chosen such that $xy\neq-4$ for any $(x,y)\in[a,b]^2$ . If in particular $ab>-4$ , then we can write

$=\dfrac1a\ln\dfrac{4+ab}{4+a^2}+\dfrac1b\ln\dfrac{4+ab}{4+b^2}$

and you can combine the logarithms if you like as

$=\ln\sqrt[a]{\dfrac{4+ab}{4+a^2}}+\ln\sqrt[b]{\dfrac{4+ab}{4+b^2}}$
$=\ln\left(\sqrt[a]{\dfrac{4+ab}{4+a^2}}\sqrt[b]{\dfrac{4+ab}{4+b^2}}\right)$

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3 years ago