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ch4aika [34]
3 years ago
8

How do I compare fractions 7/10 and5/12 using benchmark

Mathematics
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

first you have to find a common denominator like 60. to do this you need to multiply the first fraction by 6 and the second fraction by 5. which will give you the fractions 42/60 and 25/60. which will show you that the first fraction is greater than the second.

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Gary wants to buy a bicycle that usually sells for $74.99. The store is having a sale and all merchandise is discounted by 25%.
NeX [460]

Answer:

B. $56.24

Step-by-step explanation:

74.99 x 7.5%=5.62425

5.62425 x 10= 56.24

7 0
2 years ago
Simplify the following by removing the brackets: a (b-c)+d (b-c)<br>​
netineya [11]

Answer:

ab - ac + db - dc

Step-by-step explanation:

you distribute the variables inside the parenthesis

4 0
3 years ago
Need help asap pls and ty.
Anastasy [175]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Please answer and explain how you received that answer, these are the three problems I had trouble with out of 18
Varvara68 [4.7K]
9) x.14 = 6.26 ==> 14x=156 ==> x=11.1

12) OAB = Right Triangle : (27+24)² = 24² + AB² (Pythagoras)
 ==> 51² =24² + AB²  ==> 2,601 = 576 + AB² ==> AB² = 2,601 - 576 = 2,025
==> AB² = 2015 AND AB =√2025 = 45
Sorry I can't read the last page
5 0
3 years ago
How do you check is the equation is a extraneous solution pr not?
vampirchik [111]

We can check the equation is an extraneous solution or not by finding the solution and plugging it in the equation, to verify it.

<h3>What is the extraneous solution?</h3>

In mathematics, an extraneous solution is a solution that comes from the process of solving the problem but is not a genuine solution to the problem, such as the answer to an equation.

As we know from the definition the extraneous solution is a solution that comes from the process of solving the problem but is not a genuine solution to the problem.

Let's suppose the equation is:

\rm \sqrt{x+4} = x -2

Squaring both the sides:

x + 4 = (x - 2)²

x + 4 = x² - 4x + 4

x² -5x = 0

x = 0 or x = 5

Checking for the solution by plugging in the equation;

\rm \sqrt{5+4} = 5 -2

3 = 3

\rm \sqrt{0+4} = 0 -2

2 ≠ -2

The solution x = 0 shows extraneous solution.

Thus, we can check the equation is an extraneous solution or not by finding the solution and plugging it in the equation, to verify it.

Learn more about the extraneous solution here:

brainly.com/question/14054707

#SPJ1

6 0
2 years ago
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