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3 years ago

8

How can you use a drawing to construct an argument? (ONLY IF YOU KNOW)

1 answer:

mamaluj [8]3 years ago

7 0

Draw something personally offensive to the viewer (nothing inappropriate). Or you could simply add the subjects of your argument into the drawing.

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(1) Area: 35 y² +13y-12

Fiesta28 [93]

Answer:

35y2+13y−12

=35y2+28y−15y−12

=7y(5y+4)−3(5y+4)

=(7y−3)(5y+4)

Therefore, the possible length and breadth are 7y-3 units and 5y+4 units

6 0

1 year 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

Which statement about the scenario represented in the table is true? Assume time is the independent variable

Verizon [17]

Answer:B

Step-by-step explanation:

5 0

3 years ago

34x+5−12x=3 what is the value of x?

serg [7]

Answer:

$\large\boxed{x=-\dfrac{1}{11}}$

Step-by-step explanation:

$34x+5-12x=3\qquad\text{subtraxct 5 from both sides}\\\\34x-12x=3-5\\\\22x=-2\qquad\text{divide both sides by 22}\\\\x=\dfrac{-2}{22}\\\\x=-\dfrac{2:2}{22:2}\\\\x=-\dfrac{1}{11}$

8 0

3 years ago

Read 2 more answers

I need help on this please

Romashka [77]

Answer:

with what

Step-by-step explanation:

4 0

3 years ago