Ask question

Ask question

All categories

2 years ago

9

Complete this word[tex] _ r _ch_i_ [\tex]

1 answer:

Anna007 [38]2 years ago

3 0

Answer:

Arachnid is my guess.

You might be interested in

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

2 years ago

Evaluate 4 - 6 + 9 - 3

omeli [17]

$4 - 6 + 9 - 3$

$= 4$

3 0

3 years ago

What is the square of 0.81?

uysha [10]

Answer:

0.6561

Step-by-step explanation:

i think i could be wrong if am right brainliest

7 0

2 years ago

Read 2 more answers

What is x? 3x+43=100

jenyasd209 [6]

Answer:

x = 19

Step-by-step explanation:

Solve for x:

3 x + 43 = 100

Subtract 43 from both sides:

3 x + (43 - 43) = 100 - 43

43 - 43 = 0:

3 x = 100 - 43

100 - 43 = 57:

3 x = 57

Divide both sides of 3 x = 57 by 3:

(3 x)/3 = 57/3

3/3 = 1:

x = 57/3

The gcd of 57 and 3 is 3, so 57/3 = (3×19)/(3×1) = 3/3×19 = 19:

Answer: x = 19

8 0

2 years ago

Read 2 more answers

Fins the zeroes of the quadratic equation : 4√ 5 x^ 2 - 24x - 9√ 5

liubo4ka [24]

Answer:

Step-by-step explanation:

Hello,

First of all, we know that the solution of the following equation

$ax^2+bx+c=0$

are

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \ \text{ when } \Delta=b^2-4ac \geq 0$

<em>I would suggest that you try to apply this formula first and check the solution only after you try.</em>

Let's apply it in this case, we have:

$a=4\sqrt{5} \\ \\ b=-24 \\ \\ c= -9\sqrt{5}$

$\Delta=b^2-4ac=24^2+4*9*5=1296 \\ \\ \sqrt{\Delta}=36 \ \text{ and the solutions are } \\ \\ \\ x_1=\dfrac{24-36}{8\sqrt{5}}=\dfrac{-12}{8\sqrt{5}}=\boxed{\dfrac{-3}{2\sqrt{5}}} \\ \\x_2=\dfrac{24+36}{8\sqrt{5}}=\dfrac{60}{8\sqrt{5}}=\boxed{\dfrac{15}{2\sqrt{5}}}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0

3 years ago