Help with question # 16. Please and thank you.
1 answer:
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The second leftover expression is not o(a+b). It is 6(a + b). I have attached the correct question to depict that.
Answer:
The equivalent expressions are;
8a + 2 and 6a + 6b
Step-by-step explanation:
The two leftover expressions are given as;
2(4x + 1) and 6(a + b)
In algebra, equivalent expressions are simply those expressions which when simplified, give the same resulting expression as the initial one.
Thus simply means expanding or collecting like times to make it clearer.
Now, in our question, like terms have already been collected. This means that to find an equivalent expression, we will just expand the bracket.
Thus;
2(4x + 1) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;
8x + 2
Similarly,
6(a + b) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;
6a + 6b
Thus;
The equivalent expressions are;
8a + 2 and 6a + 6b
The vertex-form equation is
y = a(x+1)² -16
Putting in the y-intercept values, we have
-15 = a(0+1)² -16
1 = a . . . . . . . . . . . add 16
Then the x-intercepts can be found where y=0.
0 = (x+1)² -16
16 = (x+1)²
±4 = x+1
x = -1 ± 4 = {-5, 3}
y = a(x+1)² -16
Putting in the y-intercept values, we have
-15 = a(0+1)² -16
1 = a . . . . . . . . . . . add 16
Then the x-intercepts can be found where y=0.
0 = (x+1)² -16
16 = (x+1)²
±4 = x+1
x = -1 ± 4 = {-5, 3}