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3 years ago

7

What can be useful when finding the value of a variable

1 answer:

jeyben [28]3 years ago

8 0

Writing an equation would be most helpful, but depending on the situation drawing a diagram or reading a table could work better.

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<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-%2022x%20%2B121" id="TexFormula1" title="x^{2} - 22x +121" alt="x^{2} - 22x +12

cupoosta [38]

x2 - 22x + 121

= x2 - 11x - 11x +121

= x(x-11) - 11 ( x -11)

= (x-11) (x-11)

6 0

3 years ago

Solve on [0,2pi). Answer in order from smallest to biggest<br> cos x sin x = sin x

Musya8 [376]

Answer:

0, π

Step-by-step explanation:

cos x sin x = sin x

cos x sin x - sin x = 0

sin x (cos x - 1) = 0

sin x = 0, x = 0, π

cos x - 1 = 0

cos x = 1, x = 0

So answers: 0, π

3 0

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

Show your work .4+6 2/5=

Temka [501]

Answer:

52/5 or 10.2

Step-by-step explanation:

6 0

3 years ago

Each statement describes a transformation of the graph of y = x. Which statement correctly describes the graph of y = x + 9? A.

leva [86]

Answer:

C. It is the graph of y = x translated 9 units to the left.

Step-by-step explanation:

C. It is the graph of y = x translated 9 units to the left.

7 0

3 years ago