To solve this problem, let us first lay out all the
factors of each number.
48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
56 : 1, 2, 4, 7, 8, 14, 28, 56
The greatest number of bouquets that can be made would be
equal to the greatest common factor of the two numbers. In this case it would
be 8.
Answer:
<span>8 bouquets</span>
We have to choose the correct answer for the center of the circumscribed circle of a triangle. The center of the circumscribed circle of a triangle is where the perpendicular bisectors of a triangle intersects. In this case P1P2 and Q1Q2 are perpendicular bisectors of sides AB and BC, respectively and they intersect at point P. S is the point where the angle bisectors intersect ( it is the center of the inscribed circle ). Answer: <span>P.</span>
Answer:
10 + 0.8 + 0.006 = 10.806
Step-by-step explanation:
Pick any two decimal numbers. The third one will be the difference between 10.806 and the sum of the two numbers you picked.
For example, choose 938.2 and 3.0055. Their sum is 941.2055. Then the third number that will make the sum be 10.806 is ...
10.806 -941.2055 = -930.3995
So, your three numbers could be ...
{938.2, 3.0055, -930.3995}
Answer:
Step-by-step explanation:
Given is a differential equation of III order,

The characteristic equation would be cubic as

By trial and error, we find that

Thus m=2 is one solution
Since given that
is one solution we get
m = -4+i and hence other root is conjugate 
Hence general solution would be

Answer:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
Step-by-step explanation:
Equation I: 4x − 5y = 4
Equation II: 2x + 3y = 2
These equation can only be solved by Elimination method
Where to Eliminate x :
We Multiply Equation I by a coefficient of x in Equation II and Equation II by the coefficient of x in Equation I
Hence:
Equation I: 4x − 5y = 4 × 2
Equation II: 2x + 3y = 2 × 4
8x - 10y = 20
8x +12y = 6
Therefore, the valid reason using the given solution method to solve the system of equations shown is:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.