Answer:
x = 24/5
Step-by-step explanation:
16+5x=40
5x=40-16
5x= 24
x= 24/5
Answer:
You have to find the volume!! Length x Width x Height
Volume would be the space inside! You did not attach a picture though nor numbers ??
Step-by-step explanation:
Answer:
By examining the line plot, we find the number with the maximum number of crosses or occurrences. This gives the mode of the data set. We find the smallest and the largest number from the data set and find their difference and this difference is the range of the given data set.
Answer:
Step-by-step explanation:
Equate the two opposite Angles
7x - 31 = 5x + 13 Subtract 5x from both sides
7x - 5x - 31 = 13 Combine
2x - 31 = 13 Add 31 to both sides
2x = 13 + 31 Combine the right
2x = 44 Divide by 2
x = 44/2
x = 22
1)
x^2 + 4x - 5
y = --------------------
3x^2 - 12
Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite
=> 3x^2 - 12 = 0 => 3x^2 = 12
=> x^2 = 12 / 3 = 4
=> x = +/-2
Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) = ∞
Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞
Limit of y when x -> - 2(+) = +∞
Limit of y when x -> - 2(-) = -
=> x = 2 and x = - 2 are a vertical asymptotes
Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite
Limi of y when x -> - ∞ = 1/3
Lim of y when x -> +∞ = 1/3
=> Horizontal asymptote y = 1/3
x - intercept => y = 0
=>
x^2 + 4x - 5
0 = ------------------ => x ^2 + 4x - 5 = 0
3x^2 - 12
Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1
=> x-intercepts x = - 5 and x = 1
Domain: all the real values except x = 2 and x = - 2
2) y = - 2 / (x - 4) - 1
using the same criteria you get:
Vertical asymptote: x = 4
Horizontal asymptote: y = - 1
Domain:all the real values except x = 4
Range: all the real values except y = - 1
3)
x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)
factor x^2 - 3x - 10 => (x - 5)(x + 2)
Multiply both sides by (x - 5) (x + 2)
=> x(x - 5) + 7( x + 2) = 14
=> x^2 - 5x + 7x + 14 = 14
=> x^2 + 2x = 0
=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0
Answer: x = 0