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Luden [163]
3 years ago
7

How many centimeters are in 6.8 meters?

Mathematics
1 answer:
sesenic [268]3 years ago
7 0

Answer:

680 cm

Step-by-step explanation:

There are 100 cm in 1 m, so multiply 6.8*10, which equals 680

hope this helps :)

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HELP ME ON THIS PLSS Kathy swims at least 6 laps everyday. Write an inequality to show how long Kathy swims each day.
mina [271]

Answer:

x ≥ 6.

Step-by-step explanation:

Given : Kathy swims at least 6 laps every day.

To find: Write an inequality to show how long Kathy swims each day.

Solution : We have given that Kathy swims at least 6 laps every day.

We can see from given statement Kathy can swim 6 laps or greater then 6 laps in a day.

Then we use greater than or equal to sign  ( for at least)

Since x ≥ 6.

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2 years ago
Please Answerrr! I need help ASAPP!
MArishka [77]

80

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
You weigh six packages and find the weights to be 34, 24, 74, 29, 69, and 64 ounces. If you include a package that weighs 154 ou
saw5 [17]

Answer:

C. the mean increases more

Step-by-step explanation:

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Felicity's dog eats no more than two cups of dog food per day. Felicity's dog eats at least one-quarter cup more than one-half o
horrorfan [7]

Answer:

\frac{1}{4}+\frac{m}{2}   \leq x\leq 2

x denotes amount of food that Felicity's dog eats

Step-by-step explanation:

Given:

Felicity's dog eats no more than two cups of dog food per day.

Felicity's dog eats at least one-quarter cup more than one-half of the amount Martin's dog eats.

The amount of food that Martin's dog eats is represented by using m

To find: the inequality that represents the situation

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Amount of food that Martin's dog eats = m

Amount of food that Felicity's dog eats ≤ 2

Also,

Amount of food that Felicity's dog eats \geq \frac{1}{4}+\frac{m}{2}

Therefore,

\frac{1}{4}+\frac{m}{2}\leq Amount of food that Felicity's dog eats ≤ 2

Let x denotes amount of food that Felicity's dog eats.

\frac{1}{4}+\frac{m}{2} ≤ \frac{1}{4}+\frac{m}{2}   \leq x\leq 2

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3 years ago
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