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3 years ago

The sum of two numbers is

2 answers:

5 0

Let the numbers be x and y

x + y = 33
-
x + 4y = 108
__________
3y = 75
__________

3y = 75

y = 75/3

y = 25

x + y = 33
x + 25 = 33
x = 33 - 25 =8

The numbers are 8 and 25.

bekas [8.4K]3 years ago

3 0

Okay you have 2 sum unknow lets call them X+Y

so X+Y=33, 4X=108

Get X by it self then plug it in 4X=108

X=-Y+33
4(-y+33)= 108
-4y+132=108 Then subtract the 132
-4y=-24 Y=6 then to find X subtract 33-6=27
Y=6 X=27 if you check it works out

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Step-by-step explanation:

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3 years ago

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8 0

2 years ago

Please show your work.

Nastasia [14]

Answer:

1. 8.19, 2. 25.96, 3. 3.95

Step-by-step explanation:

use soh cah toa

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multiply both sides by 28

28cos73=x=8.186

2. sin is opposite over hypotenuse

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8 0

3 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago