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3 years ago

14

A recipe calls for 3 2/3 cups of flour. Earl used 7 1/3 cups. By how much did he increase the recipe

2 answers:

kolezko [41]3 years ago

8 0

3 2/3 =3.66
7 1/3 =7.33
7.33-3.66=3.77

Fynjy0 [20]3 years ago

6 0

Answer:

Earl doubled the number of recipies due to amount of flour.

Step-by-step explanation:

A recipe needs $3\,\frac{2}{3}$ cups of flour $\left(\frac{11}{3}\right)$ , whereas Earl made use of $7 \,\frac{1}{3}$ cups $\left(\frac{22}{3} \right)$ . The number of equivalent recipies is determined by simple rule of three:

$x = \frac{\frac{22}{3} }{\frac{11}{3} }$

$x = 2\,recipes$

Earl doubled the number of recipies due to amount of flour.

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Step-by-step explanation:

The explicit form for an arithmetic sequence is:

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where $a_1$ is the first term and $d$ is the common difference.

The common difference here is 2 because the y's are going up by 2 while the x's are going up by 1.

Yes, the common difference is the slope.

So we have d=2.

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x is n.

y is $a_n$ .

So the answer is true.

----

You can also verify by plugging in numbers for n and see if you get the outputs mentioned in the pairs given:

Let n=1:

$a_1=9.5+2(1-1)$

$a_1=9.5+2(0)$

$a_1=9.5+0$

$a_1=9.5$

Let n=2:

$a_2=9.5+2(2-1)$

$a_2=9.5+2(1)$

$a_2=9.5+2$

$a_2=11.5$

Let n=3:

$a_3=9.5+2(3-1)$

$a_3=9.5+2(2)$

$a_3=9.5+4$

$a_3=13.5$

Let n=4:

$a_4=9.5+2(4-1)$

$a_4=9.5+2(3)$

$a_4=9.5+6$

$a_4=15.5$

Let n=5:

$a_5=9.5+2(5-1)$

$a_5=9.5+2(4)$

$a_5=9.5+8$

$a_5=17.5$

Let n=6:

$a_6=9.5+2(6-1)$

$a_6=9.5+2(5)$

$a_6=9.5+10$

$a_6=19.5$

We have confirmed that we get all 6 of the mentioned points using the equation they gave.

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