Answer:
a - 2
Step-by-step explanation:
Perimeter = side 1 + side 2 + side 3
So
6a + 3 = 2(a + 3) + (3a - 1) + side 3
side 3 = (6a + 3) - (2a + 6) - (3a - 1)
side 3 = 6a - 2a - 3a + 3 - 6 + 1 = a - 2
The other end point is 0,4
Answer:
C. Kalena made a mistake in Step 3. The justification should state: -x²
+ x²
Step-by-step explanation:
Given the function x(x - 1)(x + 1) = x3 - X
To justify kelena proof
We will need to show if the two equations are equal.
Starting from the RHS with function x³-x
First we will factor out the common factor which is 'x' to have;
x(x²-1)
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;
a²-b² = (a+b)(a-b) hence;
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Factorising x(x+1) gives x²+x, therefore
x(x+1)(x-1) = (x²+x)(x-1)
(x²+x)(x-1) = x³-x²+x²-x
The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²
4s+7a=861
s+a=168
This can be solved using either elimination or substitution. I am going to use substitution.
Solve s+a=168 for s
s=168-a
Replace 168-a for s in 4s+7a=861
4(168-a)+7a=861
672-4a+7a=861
Solve for a
672+3a=861
3a=189
a=63
Substitute 63 for a in s=168-a
s=168-63=105
So, s=105 student tickets and a=63 adult tickets
