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jolli1 [7]
3 years ago
15

A. 4 packs of pencils and 3 packs of erasers B. 4 packs of pencils and 5 packs of erasers C. 5 packs of pencils and 3 packs of e

rasers D. 10 packs of pencils and 6 packs of erasers

Mathematics
1 answer:
amid [387]3 years ago
3 0
C. 5 packs of pencils and 3 packs of erasers. it is c because no matter what 20(the pack of erasers) is multiplied by it will always end in a zero, therefore you would have to multiply 12 by a number that would make it a multiple of 20 as well which happens to be 5. 5×12=60 and 3×20=60
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A new Bruno's Bakery uses 4/9 of a barrel of raisins each day. Cosmo's Bakery uses 6/5 as much raisins as the Bruno's Bakery. Ho
navik [9.2K]
So we know that the Bruno's BAkery uses 4/9 barrels and that Cosmo's Bakery uses 6/5 as much. So we have to multiply 4/9 * 6/5 which is equal to 0.53 for one day.
Since they are asking for the 3/7 of the week we multiply 3/7 *0.53.
This is equal to 0.23 or 23/100
:3
3 0
3 years ago
The perimeter of the figure to the right, made up of identical squares, is equal to 56 cm. What is the area of the figure?
Mamont248 [21]

The area of the square will be 196cm².

<h3>How to calculate the area?</h3>

It should be noted that the perimeter of a square is the addition of all its sides. Therefore, the length of the side will be:

= 56/4

= 14

Therefore, the area of the figure will be:

= 14²

= 14 × 14

= 196cm²

Learn more about area on:

brainly.com/question/25292087

#SPJ1

4 0
2 years ago
The difference between the square roots of a number is 30. what is the answer?
Sveta_85 [38]
Since 30 is not a square root you would use the closest square roots √25 and √36 which makes your answer 5 and 6
6 0
3 years ago
Find the volume of the region bounded above by the elliptical paraboloid zequals=9 minus 4 x squared minus 5 y squared9−4x2−5y2
Virty [35]

Answer:

\int\limits^1_0 \int^1_0{z} \, dA

Step-by-step explanation:

\int\limits^1_0\int^1_0 {9-4x^2-5y^2dy} \, dx

\int\limits^1_0 {9y-4x^2y-\frac{5y^3}{3}|_0 ^1 } \, dx

\int\limits^1_0 {22/3 -4x^2} \, dx

You finish the leftover.

7 0
3 years ago
You roll two standard number cubes. What is the probability that the sum is odd, given than one of the number cubes shows a 1?
Virty [35]
 It's not specified whether 1 is the 1st or 2nd roll: HOWER:

The 1st Roll is "1": P(odd sum/the 1st Roll is 1)
What is the sample space of all numbers starting with "1":
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),} = 6
the couple of add sum=(1,2), (1,4), (1,6), =3
P(odd sum/ 1st is 1) = 3/6 =1/2
or in applying the formula:

P(odd sum/the 1st Roll is 1) =P(odd sum ∩ 1) / P(getting "1") it will give the same probability = 1/2

NOW if the 2nd Roll is "1", it 's still 1/2

6 0
3 years ago
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