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REY [17]
3 years ago
8

Can you help me please??

Mathematics
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

205 grams

Step-by-step explanation:

Add up the individual weights of each ingredient of the snack...

60 + 75 + 40 + 30 = 205 grams

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Multiply x(x 5) how to find the answer
PilotLPTM [1.2K]
I hope this helps you



x.x^5



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x^6
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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
4 years ago
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xz_007 [3.2K]
12500 cause 5000 goes into 12500 twice and it would be 24$ for 10000 sqft and then another 6 so 30$ cause 2500 is half of 5000
3 0
3 years ago
The volume V (in cubic feet) of a right cylinder with a height of 6 feet and radius r (in feet) is given by V=6πr2 . Solve the f
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Answer:

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Step-by-step explanation:

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r=8 ft

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What is 3 less than -2
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-2 - 3 = -5
The answer is negative 5.
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