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1 year ago

7

With the given information, which of these conclusions is reasonable?

1 answer:

Schach [20]1 year ago

7 0

To which character in Seedfolks does Paul Fleischman assign the most personal perspective on gardening?Kim Wendell

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Find the value of x.<br> 3<br> X<br> x = [?]

DedPeter [7]

I think it would be 2 because it looks like 6 is the total and 3 is a side number so if you were to multiply 3 times 2 it would be 6. I hope this is correct!

3 0

2 years ago

Which graph best represents a functionwith zeros of -3, 0, and 2?

torisob [31]

Answer:

0

Step-by-step explanation:

5 0

2 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

Is 26 divisible by 2

sergij07 [2.7K]

Yes because 26 is an even number

3 0

3 years ago

Read 2 more answers

Quadratic equations and complex numbers PLEASE HELPPPP ASAP

kondor19780726 [428]

we are given

$27x^3-1=0$

we can also write it as

$(3x)^3-(1)^3=0$

now, we can use factor formula

$a^3-b^3=(a-b)(a^2+ab+b^2)$

we can use above formula

we get

$(3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)$

now, we can simplify it

$27x^3-1=(3x-1)(9x^2+3x+1)$

now, we can set it to 0

$27x^3-1=(3x-1)(9x^2+3x+1)=0$

and then we can solve for x

$(3x-1)(9x^2+3x+1)=0$

$3x-1=0$

$x=\frac{1}{3}$

$9x^2+3x+1=0$

now, we can use quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

now, we can plug values

and we get

$x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}$

$x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}$

So, we will get solution as

$x=\frac{1}{3},\:x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}$ ...............Answer

5 0

2 years ago