Answer:
First question:
The graph of has a vertical asymptote at x = and a horizontal asymptote at y =
Second question:
The graph of equation has a horizontal asymptote at y = -3 ⇒ C
Step-by-step explanation:
The vertical asymptotes will occur at the values of x for which make the denominator is equal to zero
The horizontal asymptotes will occur if:
First question:
∵
- To find the vertical asymptote equate the denominator by 0
to find the value of x
∵ The denominator is 2 - 3x
∴ 2 - 3x = 0
- Add 3x to both sides
∴ 2 = 3x
- Divide both sides by 3
∴ = x
∴ The graph has a vertical asymptote at x =
To find the horizontal asymptote look at the highest degree of x in both numerator and denominator
∵ The denominator and the numerator has the same degree of x
- Divide the coefficient of x of the numerator and denominator
∵ The coefficient of x in the numerator is -2
∵ The coefficient of x in the denominator is -3
∵ -2 ÷ -3 =
∴ The graph has a horizontal asymptote at y =
The graph of has a vertical asymptote at x = and a horizontal asymptote at y =
Second question:
The graph has a horizontal asymptote at y = -3
means the numerator and the denominator has same highest degree and the coefficient of the highest degree in the numerator divided by the coefficient of the highest degree in the denominator equal to -3
∵ In answer A the quotient is 1 because x up and down have
coefficient 1
∵ In answer B the quotient is because the coefficient of x
up is 1 and down is -3
∵ In answer D the quotient is -1 because the coefficient of x
up is 3 and down is -3
∵ In answer C the quotient is -3 because the coefficient of x up
is -3 and down is 1
∴ The graph of equation has a horizontal asymptote at y = -3
As per the concept of unitary method, Aaron collected 7 shells.
Let's assume that Aaron initially collected "A" shells, and Maria initially collected "M" shells.
According to the first piece of information, Maria had three times as many shells as Aaron: M = 3A.
Now, according to the second piece of information, if Aaron had found 8 more shells and Maria had found 6 fewer shells, their final shell counts would have been the same. So, Aaron's final count would be A + 8, and Maria's final count would be M - 6.
Given that the final counts are equal, we can set up an equation:
A + 8 = M - 6.
Now, we can substitute the value of M from the first piece of information (M = 3A) into the equation above:
A + 8 = 3A - 6.
Next, let's solve for A:
8 + 6 = 3A - A,
14 = 2A.
Finally, we can find the value of A by dividing both sides of the equation by 2:
A = 14/2,
A = 7.
So, Aaron collected 7 shells.
To know more about unitary method here
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B. 6n^2 – 12n + 8
C. 6n^2 – 16n + 24
D. 4n^2
E. 24n – 24
Answer:
B.
Step-by-step explanation:
Given,
The number of smaller cubes =
So, the number of cubes that have no coloured faces. = ,
Note : If a cube painted outside having side n is split into n³ cubes, then the volume volume that is not painted = (n-2)³
Thus, the remaining cubes that have been painted red on at least one of their faces
= Total cubes - cubes with no painted face
Hence, OPTION B is correct.
The maximum area that can be roped off with 200 feet of rope is 2500 square feet by making the roped off area a square.
The question deals with the optimization of area given a fixed perimeter, which involves the principles of geometry and algebra. Since the area needs to be roped off is a rectangle, and you have 200 feet of rope, your rectangle will have dimensions length (L) and width (W) such that 2L + 2W = 200.
To maximize the area of a rectangle given a fixed perimeter, the rectangle should be a square. So, for a maximum area, the length and width should be equal. Thus, each dimension (length and width) would be 200/4 = 50 feet.
Finally, to find the maximum area, we multiply the length by the width: 50 feet * 50 feet = 2500 square feet. So, the maximum area that they can rope off with 200 feet of rope is 2500 square feet.
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4m^3p, 9mp^4, 18m^4p^2
Answer:
mp3 your welcome
Step-by-step explanation:
smile all day
Answer:
x=20-21
x= -1
hope helps... :)