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3 years ago

7

Mon wants to make 5 lbs of the sugar syrup. How much water and how much sugar does he need to make 75% syrup?

1 answer:

Julli [10]3 years ago

7 0

Answer:

He needs 3.75 lbs of sugar and 1.25 lbs of water.

Step-by-step explanation:

Let x lbs be the amount of sugar in the syrup. Then 5-x lbs is the amount of water in this syrup.

Note

5 lbs - 100%

x lbs - 75%

Write a proportion:

$\dfrac{5}{x}=\dfrac{100}{75}$

Cross multiply:

$100x=5\cdot 75\\ \\100x=375\\ \\x=\dfrac{375}{100}=3.75$

So, he needs 3.75 lbs of sugar and 5 - 3.75 = 1.25 lbs of water.

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Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

<em /> $S_{1893} =5632.98$ <em />

<em />

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

<em /> $S_{1893} =5632.98$ <em> ----- approximated</em>

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