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2 years ago

12

What would this be? Thanks for helping!

2 answers:

igor_vitrenko [27]2 years ago

6 0

Answer:

if I remember correctly it would be a triangle.

Scrat [10]2 years ago

5 0

Triangle would be the answer have a good day

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Plz help<br><img src="https://tex.z-dn.net/?f=%20" id="TexFormula1" title=" " alt=" " align="absmiddle" class="latex-formula">

grandymaker [24]

Cot x= cosx/sinx =√3
tan x= 1/ √3

cos x = √2 /2
x= 45
tan45= 1

tan x= -4/3

csc x= 13/5
1/sin x = 13/5
sin x= 5/13
x=22.62
tan x= 5/12

6 0

3 years ago

The social science that deal with the behavior and thinking of organisms <br> Yes

LUCKY_DIMON [66]

I think it's psychology.

6 0

3 years ago

Read 2 more answers

Help me this is really hard

nirvana33 [79]

1. 2n > 17

n > 7

2. n/5 ≥ 11

n ≥ 55

3. -3y ≤ -18

y ≥ 6

If these are right, please leave a thanks and a rating.

6 0

3 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

What is 29/33 in simplest form

Oduvanchick [21]

29/33 is already in simplest fraction form

29 is a prime number and cannot be simplified more

6 0

3 years ago

Read 2 more answers