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MAXImum [283]
2 years ago
6

What is the concentration of a new solution obtained by mixing 25ml of water with 125ml of a 20% salt solution in water?

Mathematics
1 answer:
expeople1 [14]2 years ago
5 0
Given that the concentration of salt in a 125 ml of salt solution is 20%, this means that the ratio of salt to water in the solution is 20 : 80 = 1 : 4.

Thus, the amount of salt in the 125 ml 20% salt solution is given by 20% of 125 = 0.2 x 125 = 25ml and the amount of water in the 125ml 20% salt solution is 125 - 25 = 100 ml.

When 25 ml of water is added to the 125 ml 20% salt solution, the new amount of water is 100 + 25 =125 and the amount of salt remain 25 ml.

Thus, the new volume of the solution is 125 + 25 = 150 ml and the proprtion of salt in the new solution is given by 25 / 150 = 0.1667

Therefore, the concentration of a new solution <span>obtained by mixing 25ml of water with 125ml of a 20% salt solution in water is 16.67%</span>
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Please help!! I need to get this done! Im begging you, brainliest and points!
Viefleur [7K]

Answer:

<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>

Step-by-step explanation:

See procedure below;

First let us make these two functions ( y = -x + 1 , y = x^2 + x - 2 ) equivalent;

- x + 1  = x^2 + x - 2,

-x - x + 1 + 2 - x^2 = 0,

- 2x + 3 - x^2 = 0,

x^2 + 2x - 3 = 0,

Now factor the simplified equation;

x^2 + 2x - 3 = 0,

( x^2 - x ) + ( 3x - 3 ) = 0,

x ( x - 1 ) + 3 ( x - 1 ) = 0,

( x - 1 )( x + 3 ) = 0,

And solve for x;

x - 1 = 0, and x + 3 = 0,

x = 1, and x = - 3

Now substitute this value of x into the two functions as to receive the y  values for each x - value;

y = - ( 1 ) + 1, <em>y = 0 for x = 1</em>,

y = ( - 3 )^2 - 3 - 2, y = 9 - 3 - 2, <em>y = 4 for x = - 3</em>,

<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>

8 0
1 year ago
The regular octagon has a perimeter of 122.4 cm. A regular octagon with a radius of 20 centimeters and a perimeter of 122.4 cent
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Step-by-step explanation:

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Answer:

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50*2=100

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Step-by-step explanation:


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2 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each equation with the operation yo
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The quadratic equations and their solutions are;

9 ± √33 /4 = 2x² - 9x + 6.

4 ± √6 /2 = 2x² - 8x + 5.

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4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

2x² - 8x + 5, here a = 2, b = -8, c = 5.                                                  x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4.     24 can also be written as 4 × 6 and √4 = 2. So                                                                                     x = 8 ± 2√6 / 2×2= 4±√6/2.

2x² - 10x + 3, here a = 2, b = -10, c = 3.                                                   x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So                                                                              x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.

2x² - 8x - 3, here a = 2, b = -8, c = -3.                                                    x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4.     88 can also be written as 4 × 22 and √4 = 2. So                                                                             x = 8 ± 2√22 / 2×2 = 4± √22/2.

2x² - 9x - 1, here a = 2, b = -9, c = -1.                                                     x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4.                                          x = 9 ± √89 / 4.

2x² - 9x + 6, here a = 2, b = -9, c = 6.                                                    x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4.                                                                             x = 9 ± √33 / 4

To match we solve the monomials.

1. -15u^3 + 5u^3

Adding

-15u^3 + 5u^3=-10u^3

2.  10u^3 +(-5u^3)

Adding

10u^3-5u^3=5u^3

3. 10u^3 + 5u^3

Adding

10u^3 + 5u^3=15u^3

4.  5u^3+ (-10u^3)

Adding

5u^3-10u^3 =-5u^3

Two separate ways to find the answers.

7 0
1 year ago
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Step-by-step explanation:

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