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Anton [14]
4 years ago
14

You need 1 1/4 cups of sugar to make 20 cookiesto make 6 cookies, you will need ___ cups of sugar.

Mathematics
1 answer:
Evgesh-ka [11]4 years ago
6 0
Use the rule of 3 simples 

hope i understand it right that there is 1 integer and 1/4 cups of sugar or can being 11/4 cups too 

1 integer 1/4 = 5/4 

5/4 cups of sugar for 20 cookies 
 x -------------------- for 6 cookies
----------------------------------------------

 x = 6*(5/4) / 20 = (30/4)/20 = 30/80 = 3/8 cups of sugar for 6 cookies 

hope helped 


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<em>Every </em><em>parabola </em><em>has </em><em>a </em><em>turning</em><em> point</em><em> </em><em>called </em><em>the </em><em>axis</em><em> of</em><em> symmetry</em><em>,</em><em>so </em><em>for </em><em>this </em><em>question</em><em> </em><em>you </em><em>just</em><em> </em><em>have </em><em>to </em><em>find </em><em>the </em><em>turning</em><em> </em><em>points</em>

<em>x </em><em>turning</em><em> </em><em>point</em><em>=</em><em> </em><em>-b/</em><em>2</em><em>a</em><em>,</em><em>and </em><em>the </em><em>y </em><em>turning</em><em> </em><em>point</em><em> </em><em>is </em><em>gotten</em><em> </em><em>by </em><em>replacing</em><em> </em><em>the </em><em>value </em><em>of </em><em>x </em><em>in </em><em>the </em><em>equation</em>

<em>y=</em><em>(</em><em>x+</em><em>1</em><em>)</em><em>(</em><em>x-4)</em>

<em> </em><em> </em><em>=</em><em>x^</em><em>2</em><em>-</em><em>3</em><em>x</em><em>-</em><em>4</em>

<em>a </em><em>is </em><em>1</em><em>,</em><em>b </em><em>is </em><em>-</em><em>3</em><em> </em><em>and </em><em>c </em><em>is </em><em>-</em><em>4</em>

<em>x=</em><em> </em><em>-</em><em>(</em><em>-</em><em>3</em><em>)</em><em>/</em><em>2</em><em>(</em><em>1</em><em>)</em>

<em> </em><em> </em><em>=</em><em>1</em><em>.</em><em>5</em>

<em>and </em><em>y=</em><em>x^</em><em>2</em><em>-</em><em>3</em><em>x</em><em>-</em><em>4</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>1</em><em>.</em><em>5</em><em>^</em><em>2</em><em>-</em><em>3</em><em>(</em><em>1</em><em>.</em><em>5</em><em>)</em><em>-</em><em>4</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>-</em><em>6</em><em>.</em><em>2</em><em>5</em>

<em>therefore</em><em> </em><em>the </em><em>axis </em><em>of</em><em> symmetry</em><em> is</em><em> (</em><em>1</em><em>.</em><em>5</em><em>,</em><em>-</em><em>6</em><em>.</em><em>2</em><em>5</em><em>)</em>

<em>I </em><em>hope </em><em>this </em><em>helps</em>

6 0
3 years ago
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