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3 years ago

The mass of the liquid below is 45.5 grams if placed in a beaker with water the liquid will....

Mathematics

2 answers:

Citrus2011 [14]3 years ago

7 0

Step-by-step explanation:

answer is you're question

Zepler [3.9K]3 years ago

3 0

Answer:

it will float!

Step-by-step explanation:

in order to find density, you must divide the mass by the volume. the volume is 56 ml, and the mass is 45.5. (45.5 / 56 = 0.8125) because we know the density of water is 1, and the density of this liquid is less than that of water, it will float because it is "lighter". hope it helped <33

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You deposit $2,200 in a bank account. Find the balance after 4 years if the account

nirvana33 [79]

Answer:

Step-by-step explanation:

2200(1+0.0225/12) raised to the power of 12(4)

2200(1.001875) raised to the power of 12(4)

2204.125 raised to the power of 12(4)

2.98771099E160

or 2.98771099 multiplied by 10 to the 160th power

3 0

3 years ago

Please help due tommorow

raketka [301]

Answer:

4 is b 0.15

i think 5 is b

Step-by-step explanation:

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3 years ago

Read 2 more answers

match the steps to find the equation of the parabola with focus (-1, -8.75) and directrix y = - 9.25.

choli [55]

Answer:

The matched options to the given problem is below:

Step1: Choose a point on the parabola

Step2: Find the distance from the focus to the point on the parabola.

Step3: Use (x, y).

Find the distance from the point on the parabola to the directrix.

Step4: Set the distance from focus to the point equal to the distance from directrix to the point.

Step5: Square both sides and simplify.

Step6: Write the equation of the parabola.

Step by step Explanation:

Given that the focus (-1,2) and directrix x=5

To find the equation of the parabola:

By using focus directrix property of parabola

Let S be a point and d be line

focus (-1,2) and directrix x=5 respectively

If P is any point on the parabola then p is equidistant from S and d

Focus S=(-1,2), d:x-5=0]

5 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

If C=(a,b,x,y) and D=(m,n,o,p) then c union d is

Ilya [14]

Step-by-step explanation:

here,

C={ a,b,x,y}

D ={ m,n,o,p}

now,

C union D ={a,b,x,y} union {m,n,o,p}

= {a,b,x,y,m,n,o,p}

therefore C union D ={ a,b,x,y,m,n,o,p}

hope u get it...

5 0

4 years ago