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kompoz [17]
3 years ago
15

What calculation can be used to find the value of p in the equation p^3 =8

Mathematics
1 answer:
blsea [12.9K]3 years ago
6 0

<u>Cube</u><u> </u><u>Root</u>

{p}^{3}  = 8 \\ p =  \sqrt[3]{8}  \\ p =  \sqrt[3]{ 2\times 2 \times 2}  \\ p = 2

<u>Formula</u>

{p}^{3}  - 8 = 0

Use the following formula.

{x}^{3}  -  {y}^{3}  = (x - y)( {x}^{2} + xy +  {y}^{2}  )

(p - 2)( {p}^{2}  + 2p + 4)

The expression p²+2p+4 has the discriminant less than 0 ( D < 0 ).

Thus, remove the expression and leave only p-2

p - 2 = 0 \\ p = 2

<u>Substitution</u>

The most obvious number that multiplies itself three times and equal 8 is 2.

Substitute p = 2

{p}^{3}  = 8 \\  {2}^{3}  = 8 \\ 2 \times 2 \times 2 = 8 \\ 8 = 8

The equation is true, thus 2 is the answer.

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kiruha [24]

Answer:

(4+8)(4) = (x) 3

Step-by-step explanation:

The picture shows two secants intersecting at a point outside the circle.  A beautiful pattern for this is:  along each secant, the product (multiply) the <em>entire</em> secant by the <em>outside</em> part is the same.

(entire secant)(outside part) = (entire secant)(ouside part)

(4+8)(4) = (x)(3)

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2 years ago
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
Does "much" mean to multiply?
erik [133]
No, but “of” does. “Much”’s meaning can change depending on the context
3 0
3 years ago
Read 2 more answers
F = 7n(d + dz) solve for z
nadya68 [22]

Answer:

z = F/7nd - 1

I hope this helps you!!!!!

3 0
3 years ago
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