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2 years ago

Melanie needs 5,000 chocolates for an event so far she has brought 2 cases of 1,125 chocolates each how many more does she need

Mathematics

2 answers:

slava [35]2 years ago

5 0

Answer:

2750 chocolates

Step-by-step explanation:

Find how much she has bought already:

1,125(2)

= 2,250

Then, subtract this from 5,000

5000 - 2250

= 2750

So, she still needs 2750

enyata [817]2 years ago

5 0

Answer:

2,750

Step-by-step explanation:

2 cases of 1,125 chocolates = 2,250 chocolates

5,000 - 2,250 = 2,750 more chocolates needed

(that's also 3 cases of chocolates needed :p)

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Answer:

Step-by-step explanation:

You can represent Morgan's scenario with the expression, 25w + 615, and you can represent Kendall's scenario with the expression, 15w + 975. To find out when they both have the same amount of money in their accounts, you set them equal to each other and solve for w:

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Anura found a can of paint that was full. She was able to paint 3 fence posts with the paint. What fraction of the can of paint

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2 years ago

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What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?

Akimi4 [234]

$\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------$

$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5$

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2 years ago

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