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Aneli [31]
3 years ago
10

An aquarium has seahorse tank that holds 3648 liters there are 32 liters of watrer per seahorse how many seahorses fit in the ta

nk if its completely full
Mathematics
1 answer:
Aneli [31]3 years ago
6 0

Answer: 114


Step-by-step explanation: Take 3648 and divide it by 32 to get 114


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What is 12% of 18? Show work
tatuchka [14]

Answer:

2.16

Step-by-step explanation:

12/100=0.12

18 · 0.12=2.16

7 0
3 years ago
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A
tatuchka [14]

Alright, lets say the measure of the 3rd angle (the smallest one) is  x degree, then measure of the 1st angle will be (x + 25) degree, and 2nd one: 3x degree. The sum of all the angles in any triangles is 180 degrees.

(x + 25) + 3x + x = 180

5x + 25 = 180 ---> 5x = 180 - 25 ---> 5x = 155 ---> x = 155/5 ---> x = 31o (third angle)

31 + 25 = 56o (first angle)

3 × 31 = 93o (second angle)

56 + 93 + 31 = 180

8 0
3 years ago
Identify the number that does not belong with the other three. explain your reasoning. -10/2 -13.4 sqrt 18 22.7
Oxana [17]
22.7
is the anwser
because irrational is repeating
7 0
3 years ago
The diameter of a circle is 10 cm what is the angle measure of an arc Pi centimeters long
vredina [299]
We know that

[length of a circumference]=2*pi*r
diameter=10 cm
r=10/2-----> 5 cm
[length of a circumference]=2*pi*5-----> 10*pi cm

if 360° (full circle) has a length of -----------> 10*pi cm
 X°---------------------------> pi cm
X=pi*360/10*pi------> 36°

the answer is 36°
8 0
3 years ago
One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r
Flura [38]

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

3 0
3 years ago
Read 2 more answers
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