Answer:
Step-by-step explanation:
Given that there are six different toys and they are to be distributed to three different children.
The restraint here is each child gets atleast one toy.
Let us consider the situation as this.
Since each child has to get atleast one toy no of ways to distribute
any 3 toys to the three children each. This can be done by selecting 3 toys from 6 in 6C3 ways and distributing in 3! ways
So 3 toys to each one in 6x5x4 =120 ways
Now remaining 3 toys can be given to any child.
Hence remaining 3 toys can be distributed in 3x3x3 =27 ways
Total no of ways
= 120(27)
= 3240
<span>Here let the quadratic equation be ax^2 + bx + c
We know that a=5 from the question.
Since the roots are 6 and 2, the quadratic equation would take the form of a product like (a1x-b1)(a2x-b2).
However, let's assume that a2=1 and b2=6,
Since a=5, a1=5, then 5x-b1=5(x-2). Solving this shows that b1=10
So, the equation is (5x-10)(x-6)</span>
It’s the first answer, both expressions equal 5 when substituting 2 for x because the expressions are equivalent.
1/2 of x +4
=1/2(2)=1.
1+4= 5
2+6=8
1/2(2)=1.
8-1=7
7-2=5
There should be a .94 increase making it be 141.94 at the end of the week.
Measure angle EFG = 15 degrees
Measure angle GFH = 15 degrees
