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3 years ago

PLZ HELP ASAP IM TIMED THX FIND AREA

Mathematics

1 answer:

s2008m [1.1K]3 years ago

3 0

Answer: B

Step-by-step explanation:

I used heron's formula.

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Vinay made a password that consists of 3 digits. none of the digits are the same. how many possible passwords did vinay choose f

zhuklara [117]

Assuming the possible digits are from 0-9, there are 10*9*8 possibilities which come out to 720

5 0

3 years ago

Read 2 more answers

Plz help with numbers 3 and 4 !!

goblinko [34]

Answer:

3. opposite

30°

4. 50° + 30° + 2x = 180°

80° + 2x = 180°

2x = 100°

4 0

2 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

Use the binomial expression (p+q)n

zvonat [6]

Answer:

{0.16807, 0.36015, 0.3087, 0.1323, 0.02835, 0.00243}

Step-by-step explanation:

The expansion of (p+q)^n for n = 5 is ...

(p+q)^5 = p^5 +5·p^4·q +10·p^3·q^2 +10·p^2·q^3 +5·p·q^4 +q^5

When the probability p=0.3 and q = 1-p = 0.7 the terms of this series correspond to the probabilities of 5, 4, 3, 2, 1, and 0 favorable outcomes out of 5 trials.

For example, p^5 = 0.3^5 = 0.00243 is the probability of 5 favorable outcomes in 5 trials where the probability of each favorable outcome is 0.3.

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The attachment shows the calculation of these numbers using a graphing calculator. It lists them in reverse order of the expansion of (p+q)^5 shown above, so that they are the probabilities of 0–5 favorable outcomes in the order 0–5.