The scale factor of the dilation is calculated by dividing the length of the line after dilation by the length of the initial line. The length of a line is calculated using the distance formula. The center of dilation is the point from which the shape scales, but we can't determine its location without more information.
To find the scale factor of a dilation in a two-dimensional plane, we divide the length of the line after dilation by the length of the initial line. The length of a line can be calculated using the distance formula which is √[(x₂ - x₁)² + (y₂ - y₁)²].
For AB, the distance will be √[(5 - 9)² + (-4 - 4)²] = √[(-4)² + (-8)²] which equals √80. For A'B', the distance will be √(3 - 6)² + (-3 - 3)² = √[(-3)² + (-6)²] which equals √45. The scale factor (k) of the dilation is the ratio of these distances, so k = √45 / √80 which simplifies to √(9/16) or 3/4.
Unfortunately, without more information, such as a fixed point in the original shape, we can't determine the exact location of the center of dilation. But, conceptually, it's the point that the shape is scaling towards or away from during the dilation.
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Answer:
The coordinates of mage of L(5, 10) will be: (11, 3)
So, the 4th (last) option is correct.
Step-by-step explanation:
As we know that Translation means how much a point is moved to the left/right and up/down.
As the x-coordinate of the image point K(7, -3) is 7 which is 6 units right to the original point J(1, 4).
Also, the y-coordinate of the image point K(7, -3) is -3 which is 7 units down to the original point J(1, 4)
so it means the translation basically implies that
(x, y) → (x + 6, y - 7)
Hence, the coordinate of the image of L(5, 10) can also be determined by
using the same rule.
(x, y) → (5 + 6, 10 - 7) = (11, 3)
Therefore, the coordinates of mage of L(5, 10) will be: (11, 3)
So, the 4th (last) option is correct.
Answer:
The answer is 15 because that absolute value for anything is 15. No matter what number
Step-by-step explanation:
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
iintervals in the form 3^???? < 3^√10 < 3^s for rational numbers ????and s.
Answer:
sequence of five intervals
(1) 3³ < <
(2) < <
(3) < <
(4) < <
(5) < <
Step-by-step explanation:
as per question given data
√10 ≈ 3.162 277 7
to find out
sequence of five intervals
solution
as we have given that √10 value that is here
√10 ≈ 3.162 277 7 ........................1
so
when we find ................2
put here √10 value in equation number 2
we get that is 32.27
so
sequence of five intervals
(1) 3³ < <
(2) < <
(3) < <
(4) < <
(5) < <