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Pani-rosa [81]
3 years ago
11

A solution of salt and water contains 70 grams of water per 140 milliliters of the solution. if 1 mole of water weighs 16 grams,

2.19 moles of water would be present in 35 milliliters of the solution. [round off your answer to two decimal places. if you answer is less than one, include the zero before the decimal point.]
Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0
Be present in 35 milliliters of the solution. [round off your answer to two decimal places. if you answer is less than one, include the zero before the decimal point.]

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When there is no equal sign, the polynomial is called an ___________.
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The answer is expression
3 0
2 years ago
Work out the value of 4bc-b when b=-3 and c=2
Pavel [41]

Answer:

-18

Step-by-step explanation:

4bc-b

Let b = -3 and c=2

4 * (-3)* 2 - (-3)

-24 +6

-18

7 0
4 years ago
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The surface area of a right triangular prism is 228 square inches. The base is a right triangle with a base height of 6 inches a
umka2103 [35]
Triangle area = 6 * 8 / 2 = 24 square inches
Prism Surface Area = 2 * triangle area + height * (side a + side b + side c)
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4 0
3 years ago
A particular shoe store stocks two styles of walking shoes. Style A sells for $66.95 and Style B sells for $84.95. At the end of
Gnom [1K]

Answer:

Style A shoes sold : <u>152</u>

Style B shoes sold : <u>88</u>

Step-by-step explanation:

Let :

  • Style A shoes = x
  • Style B shoes = y

Forming equations :

  1. x = 2y - 24
  2. 66.95x + 84.95y = 17,652

Substitute the value of x from 1 in 2.

  • 66.95 (2y - 24) + 84.95y = 17,652
  • 133.90y - 1606.80 + 84.95y = 17,652
  • 218.85y = 19258.80
  • <u>y = 88</u>

<u />

Finding x :

  • x = 2(88) - 24
  • x = 176 - 24
  • <u>x = 152</u>

<u />

Solution :

  • Style A shoes sold : <u>152</u>
  • Style B shoes sold : <u>88</u>
8 0
2 years ago
50 points
Artemon [7]

Answer:

difference\hspace{0.2cm} between\hspace{0.2cm} the\hspace{0.2cm} two\hspace{0.2cm} areas\hspace{0.2cm} =\hspace{0.2cm} 9(2\sqrt{3} -\pi)

Step-by-step explanation:

i) area of circle = \pi \times r^{2}  = 9\pi \hspace{0.1cm} where \hspace{0.1cm}r = 3.

ii) therefore side of hexagon, a = 2\sqrt{3}

iii) therefore area of hexagon, A = \frac{3\sqrt{3} }{2} a^{2}  =  \frac{3\sqrt{3} }{2} \times 12 = 18\sqrt{3}

iv) difference between the two areas in iii) and i)  =9(2\sqrt{3} -\pi)

3 0
3 years ago
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