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ankoles [38]
3 years ago
12

For every 1 litre of water used to make a medicine, 199ml of sucrose and 271ml of saline solution are used. Express the amount o

f water, sucrose and saline solution needed as a ratio in its simplest form.
Mathematics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

Step-by-step explanation:

water = 1 litre = 1000 ml

water : sucrose : saline = 1000 : 199 : 271

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Why can't <img src="https://tex.z-dn.net/?f=x%5E2-5x-1" id="TexFormula1" title="x^2-5x-1" alt="x^2-5x-1" align="absmiddle" class
7nadin3 [17]

Answer:

Step-by-step explanation:

Discussion.

Not directly. But the quadratic formula can do it. But that's not your question.

the factors you get must contain the factors for 1 which are 1 and - 1

These factors must add to - 5. There's no way that will happen with 1 and - 1 and you would be creative math if you tried to say that you could make one of the factors (x -1-1-1-1-1-1). That creates a whole new question.

5 0
1 year ago
A bicycle has a listed price of $519.95 before tax. If the sales tax rate is 9.75% , find the total cost of the bicycle with sal
jek_recluse [69]
Youre finding 9.75% of 519 you know its going to be near 50 dollars
your answer is 50.695125
or 50.70$ tax
519.95+50.7 = 570.65$ total
7 0
3 years ago
Length = 25cm, bredth =60 cm, Volume = 7500cm^3. Find height of a cuboid.
elena-14-01-66 [18.8K]

Answer:

height = 5cm

Step-by-step explanation:

Volume of the cuboid = length x breadth x height

7500 = 25 x 60 x height

height = 7500 / 1500

height = 5

8 0
3 years ago
Read 2 more answers
What is three-fourths of 1,968
lara31 [8.8K]
All you have to so is divide 1968 by 4.

1968 ÷ 4 = 496.5 = 1/4


Now multiply 496.5 by 3 =  1, 489.5 = 3/4


I hope this helps! :_
7 0
3 years ago
Read 2 more answers
Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
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