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3 years ago

6

Help please I cant get this right

1 answer:

ruslelena [56]3 years ago

8 0

if it were me I would start by inputting 3 into the equation and then trying to simplify it

$\sqrt{3^2+6*3+9}$

$\sqrt{9+18+9}$

$\sqrt{36}$

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Someone sold 20 bottles. White water for 1 dollar and the black ones for 2 dollars each. He earned 35 dollars. How many of each

aivan3 [116]

Answer:

20+1+2+35= 58

Step-by-step explanation:

7 0

2 years ago

Keelie has a triangular-shaped card. The lengths of its sides are 4.4 cm, 5.5 cm, and 6.6 cm. Is the card a right triangle?

tino4ka555 [31]

Answer:

No

Step-by-step explanation:

We can use the Pytaghorean theorem to verify it

4.4^2 + 5.5^2 = 6.6^2

19.36 + 30.25 = 43.56

49.61 = 43.56 (false)

5 0

3 years ago

−7(7+5x)+x=−3x−18 what does x =

Eduardwww [97]

Answer:

-1

Step-by-step explanation:

-49-35x+x=-3x-18

-49-34x=-3x-18

-34x+3x=-18+49

-31x=31

-31x/-31=31/-31

ansx=-1

pliz mark my answer brainliest

4 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

Will give brainliest

Andrews [41]

Answer:

31.6%

6.67%

14.3%

Step-by-step explanation:

7 0

3 years ago