Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
(-2x-3)(4x+5) = -(2x+3)(4x+5) = -(8x^2+12x+10x+15) = -8x^2 - 22x - 15
Leading coefficient is -8
Givens
Start = S = - 700
# Correct = C = 8
Value Correct = VC = 400
# Incorrect =I = 9
Value Incorrect =VI = - 600
Equation
Score = S + C*VC + I*VI Substitute
Solve
Score = -700 + 8*400 + 9(-600)
Score = -700 + 3200 - 5400
Score = - 700 - 2200
Score = - 2900 <<<<< Answer
Answer:
5x^2 + 4x + 3.
Step-by-step explanation:
While I'm a little confused on what is closed and open parenthesis/brackets, I think I understand what you mean. In (5x^2+3x-2)+(3x-2x+5), the brackets can simply be removed as there is no multiplication between them. This means we get 5x^2 + 3x - 2 + 3x - 2x + 5. Combining like variables, we get 5x^2 + 4x + 3.
Answer:
1/6^3
Step-by-step explanation:
The applicable rules of exponents are ...
a^-b = 1/a^b
(a^b)(a^c) = a^(b+c)
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Your expression can be simplified as follows:

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<em>Additional comment</em>
If you think of an exponent as signifying repeated multiplication, the rules of exponents may be easier to remember. The exponent tells you how many times the base is a factor in the product.
Consider multiplication:

Consider division:

This may help you see that a positive exponent in the denominator is equivalent to a negative exponent in the numerator (and vice versa).