Answer:
2a^4+5a^3-6a^2+19a-20
Step-by-step explanation:
(a^2+3a-4)(2a^2-a+5)
2a^4-a^3+5a^2+6a^3-3a^2+15a-8a^2+4a-20
2a^4-a^3+6a^3+5a^2-3a^2-8a^2+15a+4a-20
2a^4+5a^3+2a^2-8a^2+19a-20
2a^4+5a^3-6a^2+19a-20
Answer:
1.
<u>An extraneous solution is a root of a transformed equation that is not a root of the original equation as it was excluded from the domain of the original equation.</u>
It emerges from the process of solving the problem as a equation.
2.I begin like:
The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:
for example:
x² − 4=0
x²= 4
doing square root on both side
x = ±2
Thus, the graph will have vertical asymptotes at x = 2 and x = −2.
To find the horizontal asymptote, the degree of the numerator is one and the degree of the denominator is two.
It could be square, rectangle, parallelogram, rhombus/diamond but, trapezoid could not be.
Answer:
P = 100 centimeters (assuming you are asking for the perimeter)
Step-by-step explanation:
P = 2 (L + W) OR P = 2L + 2W where P equals the perimeter, L equals the length, and W equals the width.
Since the rectangle's length is 4 times its width and given we know the width is 10 centimeters, then the length is 40 centimeters. Now we can substitute these values in our formula, okay?
P = 2 (L + W)
P = 2 (40 + 10)
P = 80 + 20
P = 100 centimeters
Answer:
25 because 50 minus 25 is 25