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3 years ago

5m -3m + 4 (m + 2) answer

Mathematics

2 answers:

inn [45]3 years ago

8 0

6m+8 is the answer to this question

Sauron [17]3 years ago

6 0

Answer:

6m+8

Step-by-step explanation:

1) Distribute

5m - 3m + 4m + 8

2) Combine like terms

6m+8

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ad-work [718]

Answer:

d, 3 5/8

Step-by-step explanation:

3 0

3 years ago

Solve 3(2x-5)-4x+8=-1

777dan777 [17]

Answer:

Step-by-step explanation:

Step 1:

3 ( 2x - 5 ) - 4x + 8 = - 1

Step 2:

5x - 15 - 4x + 8 = - 1

Step 3:

x - 15 + 8 = - 1

Step 4:

x - 7 = - 1

Answer:

x = 6

Hope This Helps :)

7 0

3 years ago

Pls someone help me with part b and their is 5.6 acres

kramer

$0.03+$0.04=$0.15x4=$0.60

6 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

4 years ago

Please explain how to do this question ;^;

Elena-2011 [213]

I need to know which question it is though

8 0

3 years ago