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Reptile [31]
3 years ago
8

Find two numbers whose sum is 114 and the difference is 58. (written as an equation please)

Mathematics
1 answer:
diamong [38]3 years ago
5 0

⇒

x + y = 114

x - y = 58

Add the equations downward:

2x + 0y = 172

2x = 172

x = 86

x + y = 114

86 + y = 114

y = 114 - 86

y = 28

Answer: The two numbers are 28 and 86.

<em>Please mark me brainliest</em>

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Answer:

36 Hope this helps Im so sorry if this is wrong.

Step-by-step explanation:

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2 years ago
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Drag the tiles to the correct boxes to complete the pairs match each pair of rational numbers with their sum -2/a^b+2/ab^2
Alexus [3.1K]

The pair of rational numbers with their sum are matched.

<h3 /><h3>What is rational number?</h3>

The numbers in the form of p/q is called the rational number.

1. -2/a²b +2/ab² = ( -2ab² +2a²b) / a³b³

                          = 2(a-b)/a²b²

2. 1/2ab -1/4a²b² =( 4a²b² -2ab) / 8a³b³

                            = 2ab -1 /4a²b²

3. 2/a²b³ -2/ab = (2ab -2a²b³) / a³b⁴

                         = 2( 1- ab²) /a²b³

4. 1/ab³ -1/a²b = a²b - ab³ / a³b⁴

                       = (a - b²) / a²b³

Thus, all the rational numbers are matched with their sum.

Learn more about rational numbers.

brainly.com/question/17450097

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5 0
2 years ago
Pls help <br>_______________​
andriy [413]

Answer:

1/2×4 1/2= 4 1/4

Step-by-step explanation:

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3 years ago
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The product (x + 3y)(3x – y) =
Digiron [165]

Answer: 3x²-3y²+8xy

Step-by-step explanation:

When you multiply two factors, you want to use the FOIL method.

First

Outer

Inner

Last

(x+3y)(3x-y)                  [distribute by FOIL]

3x²-xy+9xy-3y²            [combine like terms]

3x²+8xy-3y²                 [rewrite]

3x²-3y²+8xy

Now, we know the product is 3x²-3y²+8xy.

4 0
3 years ago
In a class of P students, the average of test scores is 70. In another class the average test scores is 92. When scores of the t
Basile [38]

So, since averages are defined as:

\frac{\sum_{k=1}^{P} P_k}{P}=70

So, since P are the total number of elements and P_k is the P_kth student. This is saying if we sum over each student's score and divide it by the number of students, we should get P, which is true.

So, using that logic, the other class can be represented as:

\frac{\sum_{k=1}^{N} N_k}{N}=70

We can take both of these equations and multiply them by N:

\sum_{k=1}^{P} P_k=70P

\sum_{k=1}^N N_k=92N

So, if we want to find the average of this we should add both our equations then divide by P+N, which is the number of all the students.

\frac{\sum_{k=1}^{P} P_k+\sum_{k=1}^{N}N_k}{N+P}=\frac{70P+92N}{N+P}

To make this simpler we can replace our LHS with 86, since that's the average of both classes combined.

86=\frac{92N+70P}{N+P} \implies\\ 86N+86P=92N+70P \implies \\ 16P=6N \implies \\ \frac{16P}{N}=6 \implies \\ \frac{P}{N}=\frac{6}{16}

Simplified we would have P/N=3/8.

7 0
3 years ago
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