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NNADVOKAT [17]
3 years ago
11

I posted a question similar to this but I entered it wrong too many times :(

Mathematics
2 answers:
Murrr4er [49]3 years ago
8 0

umm sorry, my sister did this and I don't know how to remove it so I just add this note I am sorry so much again I am sorry...

slava [35]3 years ago
5 0

Answer:

\sqrt[4]{\frac{5}{7} }

Step-by-step explanation:

^4√5/^4√7

~Apply radical rules

^4√5/7

Best of Luck!

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3 years ago
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John buys 3 pounds of cherries and 2 pounds of bananas he pays a total of $24.95 the bananas cost $6.50 less per pound than the
Murrr4er [49]

John will pay $8.68 for the combined cost of 1 pound of banana and 1 pound of cherries.

Let: b=cost of banana per pound and c=cost of cherries per pound

Equation 1: For 3 pounds of cherries and 2 pounds of bananas, John pays a total of $24.95.

3c + 2b =$24.95

Equation 2: The cost of bananas is $6.50 less than a pound of cherries.

b= c - $6.50

We can substitute the second equation into the first one to solve for the cost of cherries per pound.

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c = $7.59

Substituting the value of c to the second equation to solve for b.

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The combined cost of 1 pound of banana and 1 pound of cherries is $1.09 + $7.59 or $8.68.

For more information regarding the system of equations, please refer to brainly.com/question/25976025.

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4 0
1 year ago
A regular hexagon has an area of 750.8 cm. The side length is 17 cm. Find the apothem.
kompoz [17]

Since, a regular hexagon has an area of 750.8 square cm and The side length is 17 cm.

We have to find the apothem of the regular hexagon.

The formula for determining the apothem of regular hexagon is s \div {{2 \tan (\frac{180}{n})}, where 's' is any side length of regular hexagon and 'n' is the number of sides of regular hexagon.

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Option B is the correct answer.

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3 years ago
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Answer:

EASY YOUR MOM.

Step-by-step explanation:

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scoundrel [369]
To find the lowest common denominator you need to know the factors of 5 and 11.
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Keep doing that until you find the lowest common number in the lists. The LCD will be 55.
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