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
Answer:
the one with pink and white at the top, and the same one with purple and blue on the bottom
Step-by-step explanation:
we have
(a+b)(3a-b)(2a+7b)
step 1
Solve
apply distributive property
(a+b)(3a-b)=3a^2-ab-3ab-b^2=3a^2-4ab-b^2
step 2
Multiply (3a^2-4ab-b^2) by (2a+7b)
6a^3+21a^2b-8a^2b-28ab^2-2ab^2-7b^3=6a^3+13a^2b-30ab^2-7b^3
answer is
<h2>6a^3+13a^2b-30ab^2-7b^3</h2>
Answer:
A. (f + g)(1) = - 9
B. (f - g)(0) = -7
C. (fg)(3) = 0
Step-by-step explanation:
A. (f + g)(1) = f(1) + g(1) f(1) = 1^2 - 9 = 1 - 9 = - 8 g(1) = 1 - 2 = - 1
= -8 -1 = -9
B. (f - g)(0) = f(0) - g(0) f(0) = 0^2 - 9 = 0 - 9 = -9 g(0) = 0 - 2 = -2
= -9 + 2 = -7
C. (fg)(3) = f(3)(g(3) f(3) = 3^2 - 9 = 9 - 9 = 0 g(3) = 3 - 2 = 1
= 0(1) = 0
Question:
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Answer:
yes
