Part a

Mathematics

2 answers:

Lunna [17]3 years ago

7 0

A) if 3.5 fudge is 18.20 then 3.5 times 2 is 7 then 18.20 times 2 is 36.40 which is how much it will cost.
B) i believe it would be 5.20x because it’s always changing at that rate

NARA [144]3 years ago

7 0

She would pay $36.40 for 7 pounds of pb fudge the constant of proportionality would be 5.2

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It's candy time! Suppose 5 friends go out candy collecting. Starting out, Ms. Gleason gave each of them 12 pieces of candy. Then

sdas [7]

Answer:

6 hours

Step-by-step explanation:

First, since they all collected 510, you divide 510 by 5, to figure out how much each person got.

You then know that each of them got 102 candies, so now, you subtract 12 from from 102, since the teacher gave them that much in the beginning.

You're left with 90, so last step is to divide 90 by 15, since they collect 15 every hour, which you get left with 6. Meaning they collected candies for 6 hours

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3 years ago

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Black_prince [1.1K]

Answer:

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3 years ago

The population of a local species of bees can be found using an infinite geometric series where a1=860 and the common ratio is 1

erma4kov [3.2K]

$a_1=860$
$a_2=\dfrac{a_1}5$
$a_3=\dfrac{a_2}5=\dfrac{a_1}{5^2}$
$\vdots$
$a_n=\dfrac{a_{n-1}}5=\dfrac{a_{n-2}}{5^2}=\cdots=\dfrac{a_1}{5^{n-1}}$

The $k$ th partial sum is

$\displaystyle S_k=\sum_{n=1}^ka_n=\sum_{n=1}^k\frac{a_1}{5^{n-1}}$
$S_k=a_1\left(1+\dfrac15+\dfrac1{5^2}+\cdots+\dfrac1{5^{k-2}}+\dfrac1{5^{k-1}}$
$\dfrac15S_k=a_1\left(\dfrac15+\dfrac1{5^2}+\dfrac1{5^3}+\cdots+\dfrac1{5^{k-1}}+\dfrac1{5^k}\right)$

$\implies S_k-\dfrac15S_k=a_1\left(1-\dfrac1{5^k}\right)$
$\dfrac45 S_k=860\left(1-\dfrac1{5^k}\right)$
$S_k=1075-\dfrac{1075}{5^k}$

As $k\to\infty$ , we're left with

$\displaystyle\sum_{n=1}^\infty a_n=\lim_{k\to\infty}\left(1075-\frac{1075}{5^k}\right)=1075$

which is the upper limit to the population of the bees.

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3 years ago

2√3+4√18 in simplest radical form

Sedaia [141]

Answer:

$2\sqrt{3} + 12\sqrt{2}$