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Verizon [17]
3 years ago
12

Andrea's rock collection has 25 fewer rocks than Claudia's collection. Andrea's collection contains 125 rocks. Which equation ca

n be used to calculate how many rocks are in Claudia's collection? Let the variable r stand for the number of rocks in Claudia's collection.
A.125 = r -25


B.125 -25 = r


C.25 - r =125


D.125 = r + 25

I need the answer ASAP
Mathematics
2 answers:
dimulka [17.4K]3 years ago
8 0
If Andrea's rock collection has 125 rocks, and has 25 less than Claudia's..how many does Claudia have?
viva [34]3 years ago
7 0
125-25=100 which is 25 fewer than 125
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Answer:

A( multiply both sides by 6)

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Is there a even number and a odd number than mutiplys a odd number if so list 3 examples
s344n2d4d5 [400]

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Even

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Well if you have an odd number like 3, 5, 7, 9, and etc multiplied by an even number like 2, 4, 6, 8, and etc you will always get an even number

2 x 3 = 6 which is even 3+3

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3 years ago
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Anna71 [15]
It’s the last option (-6m-n+3)
3 0
2 years ago
Read 2 more answers
How many pairs of number equal the sum of 12
aleksley [76]
Total numbers of whole number that has a sum of 12. 
1. 11 + 1 = 12
2. 10 + 2 = 12
3. 9 + 3 = 12
4. 8 + 4 = 12
5. 7 + 5 = 12
6. 6 + 6 = 12
7. 12 + 0= 12

so there are 7 pairs of whole numbers being added that have a sum of 12
7 0
3 years ago
Write the following sum using summation notation -2/5+3/6-4/7........11/14
horrorfan [7]

Answer:

\sum_{n=1}^{10}(-1)^n\left ( \frac{n+1}{n+4} \right )

Step-by-step explanation:

Given: \frac{-2}{5}+\frac{3}{6}-\frac{4}{7}+...+\frac{11}{14}

To write: the given expression using summation notation

Solution:

Summation notation helps to write a long sum as a single expression.

In the summation notation, the variable \sum is called the index of summation.

Let x_1,x_2,x_3,...,x_n denote a set of n numbers.

Then in summation notation,

x_1+x_2+x_3+...+x_n=\sum_{i=1}^{n}x_i

\frac{-2}{5}=(-1)^1\left ( \frac{1+1}{1+4} \right )\\\frac{3}{6}=(-1)^2\left ( \frac{2+1}{2+4} \right )\\\frac{-4}{7}=(-1)^3\left ( \frac{3+1}{3+4} \right )\\.\\.\\.\\\frac{11}{14}=(-1)^10\left ( \frac{10+1}{10+4} \right )\\\therefore \frac{-2}{5}+\frac{3}{6}-\frac{4}{7}+...+\frac{11}{14}=\sum_{n=1}^{10}(-1)^n\left ( \frac{n+1}{n+4} \right )

3 0
3 years ago
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