(8x 2 −15x)−(x 2 −27x)=ax 2 +bxleft parenthesis, 8, x, squared, minus, 15, x, right parenthesis, minus, left parenthesis, x, squ
quester [9]
Answer:
<h2>5</h2>
Step-by-step explanation:
Given the expression (8x² −15x)−(x² −27x) = ax² +bx, we are to determine the value of b-a. Before we determine the vwlue of b-a, we need to first calculate for the value of a and b from the given expression.
On expanding the left hand side of the expression we have;
= (8x² −15x)−(x² −27x)
Open the paranthesis
= 8x² −15x−x²+27x
collect the like terms
= 8x²−x²+27x −15x
= 7x²+12x
Comparing the resulting expression with ax²+bx
7x²+12x = ax²+bx
7x² = ax²
a = 7
Also;
12x = bx
b =12
The value of b - a = 12 - 7
b -a = 5
Hence the value of b-a is equivalent to 5
The domain is the scope of the x values. In this case (-2,-1) is a hole, so x>-2 is one end of the domain, while (3,3) is defined. So the domain using interval notation is (-2,3] which can also be expressed -2 < x ≤ 3, answer option 1
Your answer would be C.) 18oz
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the following Identities:
sec Ф = 1/cos Ф
cos² Ф + sin² Ф = 1
<u>Proof LHS → RHS</u>
