Mathematics: If 39 is added to a number, it will be 4 times as large as it was originally. What s the number?

Substitute , so that . The integral is then equivalent toIn general, , so .We want the substitution made above to be reversible, so that . This restricts to the interval , and over this interval we have , so we take the positive square root in order that .Then the integral becomeswhich can be computed in several ways. One method is to integrate by parts, takingso thatThen with , you have and , which follows from the Pythagorean identity. So...

If 39 is added to a number, it will be 4 times as

Ответ:

yaneth1248

22.03.2021

Mathematics

Step-by-step explanation:

let the number=x

x+39=4x

39=3x

x=39/3=13

Check: 13+39=52

4*13=52

0,0(0 оценок)

Ответ:

andrwisawesome0

21.11.2020

Mathematics

Substitute $t=\tan s$ , so that $\mathrm dt=\sec^2s\,\mathrm ds$ . The integral is then equivalent to

$\displaystyle\int\sqrt{1+(\tan s)^2}\,\sec^2s\,\mathrm ds$

$=\displaystyle\int\sqrt{1+\tan^2s}\,\sec^2s\,\mathrm ds$

$=\displaystyle\int\sqrt{\sec^2s}\,\sec^2s\,\mathrm ds$

In general, $\sqrt{x^2}=|x|$ , so $\sqrt{\sec^2s}=|\sec s|$ .

We want the substitution made above to be reversible, so that $s=\tan^{-1}t$ . This restricts to the interval $-\dfrac\pi2$ , and over this interval we have $\cos s0\implies\sec s0$ , so we take the positive square root in order that $\sqrt{\sec^2s}=+\sec s$ .

Then the integral becomes

$=\displaystyle\int\sec^3s\,\mathrm ds$

which can be computed in several ways. One method is to integrate by parts, taking

$u=\sec s\implies\mathrm du=\sec s\tan s\,\mathrm ds$

$\mathrm dv=\sec^2s\,\mathrm ds\implies v=\tan s$

so that

$\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s\tan^2s\,\mathrm ds$

$\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s(\sec^2s-1)\,\mathrm ds$

$\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int(\sec^3s-\sec s)\,\mathrm ds$

$\displaystyle2\int\sec^3s\,\mathrm ds=\sec s\tan s+\int\sec s\,\mathrm ds$

$\displaystyle\int\sec^3s\,\mathrm ds=\frac12\sec s\tan s+\frac12\ln|\sec s+\tan s|+C$

Then with $s=\tan^{-1}t$ , you have $\tan s=t$ and $\sec s=\sqrt{1+t^2}$ , which follows from the Pythagorean identity. So

$\displaystyle\int\sqrt{1+t^2}\,\mathrm dt=\frac{t\sqrt{1+t^2}+\ln|t+\sqrt{1+t^2}|}2+C$

0,0(0 оценок)

If 39 is added to a number, it will be 4 times as large as it was originally. What's the number?

Ответ:

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question