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ANTONII [103]
3 years ago
10

Help!!!!!!!!!!!!!!!!!!!!!!!!! and I'm giving brainly thxxxxx

Mathematics
1 answer:
mash [69]3 years ago
6 0

Answer:

the 3 is in the hundred thousands

Step-by-step explanation:

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WILL GIVE 20 POINTS PLEASE HELP
Mila [183]

Answer:

y = x + 6

Step-by-step explanation:

y = x + b

when x = 0, y = 6. so....

6 = (0) + b

6 = b

so, y = x + 6

7 0
3 years ago
What is the solution to the following equation?<br> x+(-21) = 8
UNO [17]
Correct answer is 13
6 0
3 years ago
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52.) Five times a number is 45
Rom4ik [11]
52).  5x=45    Divide each side of the equation by 5.

53).  -3x = 12   Divide each side by -3.

54).  x/4 = 10    Multiply each side by  4 .
 
55).  x/3 = -8    Multiply each side by  3 .

1).  x -10 = 12 .     Add  10  to each side.

3).  x + 8 = 16      Subtract  8  from each side.   

5).  5 + x = 6        Subtract  5  from each side.

7).  x - 4 = 9         Add  4  to each side.

Thank you for the 5 points.    The crust and warm water are delicious.
4 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
Simplify 3/5+(-1/4) 7/10
Vadim26 [7]

Answer:

17/40

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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