All categories

3 years ago

Amanda's gross annual salary is $62,112 what is the maximum amount of rent she can afford to pay

Mathematics

1 answer:

abruzzese [7]3 years ago

7 0

Answer is 1866 dollars

You might be interested in

Evaluate 5r-8.35s when r=12 and s=4

balu736 [363]

Answer: 26.6

Step-by-step explanation:

r = 12 5(12) = 60 | s = 4 8.35(4) = 33.4 | Now subtract both total number = 33.4 - 60 = 26.6

7 0

3 years ago

Solve for x the rest of the question is in the png

zvonat [6]

Answer:

33.18

Step-by-step explanation:

Use law of sines.

x / sin(65°) = 21 / sin(35°)

x = 21 sin(65°) / sin(35°)

x ≈ 33.18

3 0

3 years ago

Find the perimeter. Round to the nearest whole number.<br> 15 is c and 30 is a)

ziro4ka [17]

Answer:

Step-by-step explanation:

Hope this helps.

8 0

3 years ago

Ryan has a $500 bond with a 6.5% coupon. How much interest will Ryan

oee [108]

D. Is the correct answer

3 0

3 years ago

Read 2 more answers

“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