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sertanlavr [38]
3 years ago
12

Give me example of a unit rate used in real life

Mathematics
1 answer:
lana66690 [7]3 years ago
3 0

Before you begin this lesson, please print the accompanying document, Unit Rates in Everyday Life].

Have you ever been at the grocery store and stood, staring, at two different sizes of the same item wondering which one is the better deal? If so, you are not alone. A UNIT RATE could help you out when this happens and make your purchasing decision an easy one.

In this lesson, you will learn what UNIT RATES are and how to apply them in everyday comparison situations. Click the links below and complete the appropriate sections of the Unit Rates handout.

[Note: The links below were created using the Livescribe Pulse Smartpen. If you have never watched Livescribe media before, take a few minutes to watch this very brief Livescribe orientation]

<span>What is a UNIT RATE – definitionView some examples of Unit RatesSee a process to compute Unit Rates</span>
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Answer:

119 and 8/14 I think.

Step-by-step explanation:

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PLEASE HELP ASAP
Ahat [919]

Answer:

5x+32

Step-by-step explanation:

So write out as an expression

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What is the length of the altitude of the equilateral triangle below?
Mazyrski [523]

ANSWER

The length of the altitude is 3\sqrt{3} units


<u>EXPLANATION</u>

The altitude of a triangle is the vertical height of the triangle.


From the diagram the altitude is a


Method 1: We can use Pythagoras theorem to find a


6^2=a^2+3^2


6^2-3^2=a^2


36-9=a^2


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We take the square root of both sides,

\sqrt{27}=a


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Method 2 Using Trigonometry

sin(60\degree)=\frac{a}{6}

6sin(60\degree)=a


a=6\times \frac{\sqrt{3}} {2}


a=3\sqrt{3}






6 0
3 years ago
Read 2 more answers
What is the percentage… b of sugar in the syrup made of 10 kg of water and 4 kg of sugar?
Artist 52 [7]

Answer:

<h3>28 4/7 % ≈ 29%</h3>

Step-by-step explanation:

4 kg - weight of sugar

10 kg - mass of water

4kg + 10kg = 14kg - weight of the syrup

\dfrac{4kg}{14kg}=\dfrac{2}{7}

The sugar is 2/7 of the weight of the syrup.

Convert to the percentage:

\dfrac{2}{7}=\dfrac{2}{7}\cdot100\%=\dfrac{200}{7}\%=28\dfrac{4}{7}\%\aaprox29\%

6 0
3 years ago
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Given the tables for f &amp; g below, find the following: x f ( x ) g ( x ) 0 9 7 1 4 8 2 3 6 3 5 1 4 7 5 5 1 2 6 6 0 7 0 4 8 2
d1i1m1o1n [39]

\\(i)\\&#10;\text{We know that the average rate of change of a function f(x) on inerval [a,b] is}\\&#10;\\&#10;f_{avg}=\frac{f(b)-f(a)}{b-a}\\&#10;\\&#10;\text{so using this the average rate of chang of g for x on interval }[3,7] \text{ is}\\&#10;\\&#10;g_{avg}=\frac{g(7)-g(3)}{7-3}\\&#10;\\&#10;=\frac{4-1}{4}\\&#10;\\&#10;=\frac{3}{4}\\&#10;\\&#10;\text{so average rate of change of g on interval [3,7] is }\frac{3}{4}

\\(ii)\\  \text{Average rate of chang of f for x on interval }[3,7] \text{ is}\\&#10;\\&#10;f_{avg}=\frac{f(7)-f(3)}{7-3}\\&#10;\\&#10;=\frac{0-5}{4}\\&#10;\\&#10;=\frac{-5}{4}\\&#10;\\&#10;\text{so average rate of change of g on interval [3,7] is }\frac{-5}{4}

\\(iii)\\&#10;\text{Since g(4)=5, so we have}\\&#10;\\&#10;f(g(4))=f(5)\\&#10;\\&#10;\text{since the value of f is 1 at x=5. so }\\&#10;\\&#10;f(g(4))=f(5)=1\\&#10;\\&#10;\text{hence we have, }f(g(4))= 1

\\(iv)\\&#10;\text{Since f(3)=5, so we have}\\&#10;\\&#10;g(f(3))=g(5)\\&#10;\\&#10;\text{since the value of g is 2 at x=5. so }\\&#10;\\&#10;g(f(3))=g(5)=2\\&#10;\\&#10;\text{hence we have, }g(f(3))= 2

5 0
3 years ago
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