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kobusy [5.1K]
2 years ago
13

How many 5 centimeter squares are needed to completely cover a 1 meter square without overlapping?

Mathematics
2 answers:
daser333 [38]2 years ago
8 0
100 cm in 1 m
Use the equation- b=100cm/5.
b=20
Feliz [49]2 years ago
5 0
There are 100 centimeters in a meter 
100 divided by 5 is 10 
You need 10 5 centimeter squares to completely cover 1 meter square without overlapping 
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Two pounds of grapes cost 6$ what number belongs in the most right empty cell?
sdas [7]

Answer:

Table shown below

Step-by-step explanation:

Table shown below

To solve this problem let's use proportions.

If 2 pounds of grapes cost $6, half the amount will cost half the dollars, so the last row will have $3 in the price

For the second row, we know the price is $1, that is, one-sixth of the original given price. It should correspond to one-sixth of the amount of grapes or 2/6 pounds.

Simplifying the fraction, we get 1/3 or 0.33 pounds

6 0
3 years ago
The area of a rectangular swimming pool is 360 square feet. If the length of the swimming pool is 9 feet more than its width, fi
Inessa [10]

Answer:

l = 24ft

Step-by-step explanation:

A = lw

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360 = x^2 + 9x

0 = x^2 + 9x - 360

0 = (x + 24)(x - 15)

x ≠ -24 (length cannot be negative)

x = 15

l = 15 + 9

l = 24ft

7 0
3 years ago
Blocks numbered 0 through 9 are placed in a box, and a block is randomly picked.
aliya0001 [1]
<span>Blocks numbered 0 through 9 are placed in a box, and a block is randomly picked.=3/10

</span><span>The probability of picking an odd prime number is . The probability of picking a number greater than 0 that is also a perfect square is=3/10</span>
3 0
3 years ago
Read 2 more answers
Mark
Colt1911 [192]
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8 0
3 years ago
Radioactive Decay Beginning with 16 grams of a radioactive element whose half -life is 30 years ,the mass y(in grams) remaining
Flura [38]

Answer:

the equation should be corrected to fit the data of the problem.  With the corrected equation a mass of 0.5 grams remains after 150 years

Step-by-step explanation:

for the mass y( in grams)

y=23* (1/2)^(t/45), t ≥ 0.

the initial mass is at t=0 , then

y= 23 grams  → should be 16 grams

half-life from the equation = 45 years → should be 30 years

the correct equation should be

y=16*(1/2)^(t/30), t ≥ 0

then after 150 years  → t= 150

y=16*(1/2)^(150/30)= 16*(1/2)^5 = 16/32 = 0.5 grams

then a mass of 0.5 grams remains after 150 years

5 0
3 years ago
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