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marta [7]
3 years ago
5

Find the non permissible replacement of 1/-8x

Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0
-7x .........................
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Increase the number 120 by 70%
Karo-lina-s [1.5K]
Multiply 120 by .7 then take the product and add it to 120.

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3 years ago
A new operation is defined by a△b=a^2-b/b-a^2. 4△3.
dmitriy555 [2]

As the new mathematical operation is defined by a△b=a^2-b/b-a^2, the value of 4△3 using the same operation will be 4△3 = -1

As per the question statement, we are given a new mathematical operation a△b=a^2-b/b-a^2 and we are supposed to find the value of 4△3 using the same operation.

Given, a△b=a^2-b/b-a^2

now 4△3 = (4^2-3) / (3-4^2)  

4△3 = (16-3) / (3-16)        

4△3 = 13 / -13  

4△3 = -1

Hence, as the new mathematical operation is defined by a△b=a^2-b/b-a^2, the value of 4△3 using the same operation will be 4△3 = -1.

  • Mathematical operation: An operator in mathematics is often a mapping or function that transforms components of one space into elements of another.

To learn more about mathematical operation, click on the link given below:

brainly.com/question/8959976

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8 0
1 year ago
On a coordinate plane, a straight line and a parallelogram are shown. The straight line has a negative slope and has a formula o
monitta

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(1,3)

Step-by-step explanation:

7 0
3 years ago
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You will write a 5-paragraph essay explaining how you would solve the following equation:
sattari [20]

Answer:

x=0

Step-by-step explanation:

\frac{7}{3}(2x+3)+\frac{3}{4}(\frac{x}{5}-\frac{15}{2})=\frac{11}{8} <-- Given

\frac{14}{3}x+7+\frac{3}{20}x-\frac{45}{8}=\frac{11}{8} <-- Distributive Property

\frac{280}{60}x+7+\frac{9}{60}x-\frac{45}{8}=\frac{11}{8} <-- Find LCD of x-terms

\frac{289}{60}x+7-\frac{45}{8}=\frac{11}{8} <-- Combine Like Terms

\frac{289}{60}x+7=\frac{56}{8} <-- Add 45/8 to both sides

\frac{289}{60}x+7}=7 <-- Simplify Right Side

\frac{289}{60}x=0 <-- Subtract 7 on both sides

x=0 <-- Divide both sides by 289/60

3 0
2 years ago
What is the range of the function? f(x) = -2^x + 1
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{y|1 > y}, (-∞, 1); you did not specifically mention what notation you needed, so I wrote both [Internal Notation and Set Notation].

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3 years ago
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