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olganol [36]
3 years ago
15

If a recipe calls for 0.800 kg of flour , about how many ounces of flour does it need? (1lb=16 oz, 1 lb = 0.454 kg )

Mathematics
2 answers:
stiks02 [169]3 years ago
8 0
For this case we must take into account the following conversions of units:
 1lb = 16 oz
 1 lb = 0.454 kg
 Applying the conversion of units we have:
 (0.800) * (1 / 0.454) * (16/1) = 28.1938326
 Rounding off we have:
 28.2 oz
 Answer:
 
it needs 28.2 oz of flour
ch4aika [34]3 years ago
8 0

Answer:  The correct option is (C) 28.2 oz.

Step-by-step explanation:  Given that a recipe calls for 0.800 kg of flour.

We are to find the number of ounces of flour that the recipe needs.

Given that

1 lb=16 oz  and  1 lb = 0.454 kg

We will be using the UNITARY method for solving the given problem.

We have

0.454 kg = 1 lb

So, 1 kg will be equal to \dfrac{1}{0.454}~\textup{lb}.

Therefore, 0.8000 kg is equal to

\dfrac{1}{0.454}\times 0.800=1.762~\textup{lb}.

Now, 1 lb = 16 oz.

So, 1.762 lb = 16 × 1.762 = 28.19 = 28.2 oz.

Thus, the recipe needs 28.2 oz of flour.

Option (C) is CORRECT.

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Select the two values of x that are roots of this equation 3x^2 + 1 =5x
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Answer:

x = 1.434 and x=0.232

Step-by-step explanation:

To find the root of the equation stated above we need to:

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3x^{2} + 1 = 5x ⇒ 3x^{2} -5x + 1 = 0

(2) Divide the whole equation by 3

3x^{2} -5x + 1 = 0 ⇒ x^{2} -\frac{5}{3}x + \frac{1}{3}= 0

(3) Use the quadratic formula to solve the quadratic equation:

The quadratic formula states that the two solutions for a quadratic equation is given by:

\frac{-b±\sqrt{b^{2} - 4ac}}{2a} (1)

In this case, a = 1, b = -\frac{5}{3}, c= \frac{1}{3}

Substituiting a, b and c in equation (1) We get:

\frac{-\frac{5}{3}±\sqrt{(-\frac{5}{3})^{2} - 4(1)(\frac{1}{3})}}{2(1)} (1)

The two solutions are:

x = 1.434 and x=0.232

8 0
3 years ago
Jacob and Ayden work at a dry cleaners ironing shirts. Jacob can iron 25 shirts per hour, Ayden can iron 35 shirts per hour. Ayd
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Answer:

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3 years ago
A chemist wants to make 100 liters of a 44% acid solution. She has solutions that are 20% acid and 60% acid.
alexgriva [62]

the first solution is 20% acid, and say we'll be using "x" liters, so how many liters of just acid are in it?  well 20% of "x" or namely 0.2x.  Likewise for the 60% acid solution, if we had "y" liters of it, the amount of only acid in it is 0.6y.

\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&x&0.20&0.2x\\ \textit{2nd solution}&y&0.60&0.6y\\ \cline{2-4}&\\ mixture&100&0.44&44 \end{array}~\hfill \begin{cases} x+y=100\\\\ 0.2x+0.6y=44 \end{cases}

x+y=100\implies y=100-x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.2x+0.6(100-x)=44} \\\\\\ 0.2x+60-0.6x=44\implies -0.4x+60=44\implies -0.4x=-16 \\\\\\ x=\cfrac{-16}{-0.4}\implies \boxed{x=40}~\hfill \boxed{\stackrel{100-40}{y=60}}

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The answer is y=1/3x+2
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2 years ago
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