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3 years ago

A pipe takes three minutes to fill three fourths of I want to think how long would it take to fill the entire tank

2 answers:

7 0

X=total volume of the entire tank.
We can suggest the following equation:
(3/4)x=3
x=(3*4)/3
x=12/3
x=4

Answer: it would take to fill the entire tank 4 minutes

ehidna [41]3 years ago

3 0

4 minutes
3 minutes for 3/4 is 1 minuet for 1/4
plz mark me brinlest anwaser if you can :)

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Sandy bought 3/4 yard of red ribbon and 2/3 yard of white ribbon to make some hair bows.Altogether,how many yards of ribbon did

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3 years ago

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Which of the following is true about a parallelogram?

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2 years ago

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Which set of ordered pairs represents a function? {(0, 1), (1, 3), (1, 5), (2, 8)} {(0, 0), (1, 2), (2, 6), (2, 8)} {(0, 0), (0,

11111nata11111 [884]

Answer: {(0, 2), (1, 4), (2, 6), (3, 6)}

Step-by-step explanation:

For a relation to be considered a function, each x-value needs to have one corresponding y-value--it cannot have more than 1.

Since all the other sets of ordered pairs feature points with two x-values with different y-values, the set above is the only function of the provided options.

4 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

Need brief explanation about why false is correct

anyanavicka [17]

We are given the equation:

(a + b)! = a! + b!

Testing the given equation

In order to test it, we will let: a = 2 and b = 3

So, we can rewrite the equation as:

(2+3)! = 2! + 3!

5! = 2! + 3!

We know that (5! = 120) , (2! = 2) and (3! = 6):

120 = 2 + 6

We can see that LHS ≠ RHS,

So, we can say that the given equation is incorrect

6 0

2 years ago