The current Brainliest answer seems to be answering the question "Every integer is a multiple of which number?" rather than the question presented here.
We say that one number is a <em>multiple </em>of a second number if we can get to the first one by <em>counting by the second</em>. For example, 18 is a multiple of 6 because we can reach it by counting by 6's (6, 12, <em>18</em>). Note that, for any number we want to count by, we can always start our count at 0.
By 2's: 0, 2, 4, 6, 8
By 6's: 0, 6, 12, 18
By 7's: 0, 7, 14, 21
Because we can always "reach" 0 regardless of the integer we're counting by, we can say that <em>0 is a multiple of every integer</em>.
More formally, we say that some number n is a multiple of an integer x if we can find another integer y so that x · y = n. By this definition, 18 would be a multiple of 6 because 6 · 3 = 18, and 3 is an integer. We can use the property that the product of any number and 0 is 0 to say that x · 0 = 0, where x can be any integer we want. Since 0 is also an integer, this means that, by definition, 0 is a multiple of every integer.
<span>There are two approaches to translate this inquiry, to be specific:
You need to know a number which can go about as the ideal square root and also the ideal block root.
You need to know a number which is an ideal square and in addition an ideal 3D shape of a whole number.
In the primary case, the arrangement is straightforward. Any non-negative whole number is an ideal square root and in addition a flawless solid shape foundation of a bigger number.
A non-negative whole number, say 0, is the ideal square foundation of 0 and additionally an immaculate shape base of 0. This remains constant for all non-negative numbers starting from 0 i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
In the second case as well, the arrangement is straightforward however it involves a more legitimate approach than the primary choice.
A flawless square is a number which contains prime variables having powers which are a different of 2. So also, a flawless block is a number which includes prime variables having powers which are a numerous of 3.
Any number which includes prime components having powers which are a various of 6 will be the answer for your inquiry; a case of which would be 64 which is the ideal square of 8 and an ideal 3D shape of 4. For this situation, the number 64 can be spoken to as prime variables (i.e. 2^6) having powers (i.e. 6) which are a different of 6.</span>
IT IS 180 because 12*15 is one 180 and it cant go over 187.
Answer:
Area of given figure = 197.48 mi² (Approx.)
Step-by-step explanation:
Note;
Divide given figure into a semi-circle, rectangle, and square
Given:
Radius = 8 mi
Length of rectangle = 11 mi
Width of rectangle = 8 mi
Side of square = 3 mi
Find:
Area of given figure
Computation:
Area of given figure = Area of semi-circle + Area of rectangle + Area of square
Area of given figure = (π/2)(r²) + [L x B] + Side²
Area of given figure = (3.14/2)(8²) + [11 x 8] + 3²
Area of given figure = (1.57)(64) + [88] + 9
Area of given figure = 100.48 + [88] + 9
Area of given figure = 197.48 mi² (Approx.)
