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Fiesta28 [93]
2 years ago
5

50 ml of water is poured into a beaker. 50 ml of another liquid is poured into the same beaker. The beaker is left undisturbed f

or some time. After some time, it is observed that the liquid forms a clear layer on top of water. If the density of water is 1 g/cc, what could be the density of the second liquid?

Mathematics
1 answer:
Ipatiy [6.2K]2 years ago
7 0

Answer:

Well if you combined 50 ml to a beaker that already had 50 ml of water in it total you have 100 ml of water. The density of the second liquid would be the same as the first beaker but sense they are combined it would be 2 g/cc.

Step-by-step explanation:

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What is the answer to this equation
julia-pushkina [17]

Answer:

x = - \frac{494}{3} = - 164\frac{2}{3}

Step-by-step explanation:

75 + \frac{3}{8} x = 13\frac{1}{4}\\\\75 + \frac{3}{8} x = \frac{53}{4}\\\\\frac{3}{8}x = \frac{53}{ 4} - 75\\\\\frac{3}{8}x = \frac{53-300}{4}\\\\\frac{3}{8}x = - \frac{247}{4}\\\\x = -\frac{247 \times 8}{4 \times 3} \\\\x = -\frac{494}{3}

6 0
2 years ago
Explain how to solve 4|3x-1|+1=8​
denis-greek [22]

4|3x - 1| + 1 = 8

4|3x - 1| = 8 - 1

4|3x - 1| = 7

|3x - 1| = 7/4

Now lets think, we have 3x - 1 in module, so either its 7/4 or -7/4, we will have 7/4 at the end because of it, so we may have 2 solutions in this case:

3x - 1 = 7/4

and

3x - 1 = -7/4

So let's see:

3x - 1 = 7/4

3x = 7/4 + 1

3x = 11/4

x = 11/12

3x - 1 = -7/4

3x = -7/4 + 1

3x = -3/4

x = -1/4

So we have two possible answers, x = 11/2 and x = -1/4

8 0
2 years ago
PLEASE HELP ME!
cestrela7 [59]

Answer:

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4 0
2 years ago
13,22,31,...<br> Find the 31st term.<br> Find the 31st term.
Galina-37 [17]

Answer: 283

Step-by-step explanation:

To do this, it is helpful to get an equation you can use to solve any term.

This equation is:

13 + 9(n-1)

So simply plug in 31 for n to get

13+9(31-1)

=13+270

=283

5 0
2 years ago
Read 2 more answers
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
2 years ago
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