In an arithmetic sequence, the difference between consecutive terms is always the same. To find the common difference, we can subtract any term from its previous term.
First, we'll find the common difference:
Common difference = (x+12) - (5x+7)
= x+12 - 5x - 7
= -4x + 5
We can check that the common difference is the same for the other pairs of consecutive terms.
To find the 8th term, we'll use the formula:
an = a1 + (n-1)d
where an is the nth term, a1 is the first term, n is the term we want to find, and d is the common difference.
We already know that the common difference is -4x + 5.
So, to find the first term, we plug in n=1:
a1 = 5x + 7
Now we can find the 8th term:
a8 = a1 + 7d
= (5x + 7) + 7(-4x + 5)
= 5x + 7 - 28x + 35
= -23x + 42
Therefore, the 8th term of the given arithmetic sequence is -23x + 42.
Please show all your work
Find the 8th term of the arithmetic sequence 5x + 7, x +12, -3x + 17
3 answers
Find the equations of the asymptotes of the hyperbola defined by the equation shown below. If necessary, round to the nearest tenth.
−64x^2+ 9y^2- 1024x + 144y-4096 = 0
Asymptotes: y=
and y=
−64x^2+ 9y^2- 1024x + 144y-4096 = 0
Asymptotes: y=
and y=
To find the equations of the asymptotes of a hyperbola, we need to first put the given equation in standard form. The standard form of the equation of a hyperbola with center (h,k) and transverse axis parallel to the x-axis is:
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1
where a and b are the lengths of the semi-major and semi-minor axes, respectively.
To put the given equation in standard form, we need to complete the square for both the x and y terms:
-64x^2 - 1024x + 9y^2 + 144y = 4096
-64(x^2 + 16x) + 9(y^2 + 16y) = 4096 + 64*16 - 9*16
-64(x + 8)^2 + 9(y + 8)^2 = 4225
[(x + 8)^2 / (-4225/64)] - [(y + 8)^2 / (4225/9)] = 1
Comparing this to the standard form, we see that:
h = -8
k = -8
a^2 = -4225/64
b^2 = 4225/9
Since the transverse axis is parallel to the x-axis, the asymptotes will have the form:
y = ± (b/a) (x - h) + k
Plugging in the values from the standard form, we get:
y = ± (√(4225/9) / √(-4225/64)) (x + 8) - 8
y = ± (64/27) (x + 8) - 8
Simplifying and rounding to the nearest tenth, we get:
y = (2.4)x - 31.6 and y = (-2.4)x + 15.6 as the equations of the asymptotes.
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1
where a and b are the lengths of the semi-major and semi-minor axes, respectively.
To put the given equation in standard form, we need to complete the square for both the x and y terms:
-64x^2 - 1024x + 9y^2 + 144y = 4096
-64(x^2 + 16x) + 9(y^2 + 16y) = 4096 + 64*16 - 9*16
-64(x + 8)^2 + 9(y + 8)^2 = 4225
[(x + 8)^2 / (-4225/64)] - [(y + 8)^2 / (4225/9)] = 1
Comparing this to the standard form, we see that:
h = -8
k = -8
a^2 = -4225/64
b^2 = 4225/9
Since the transverse axis is parallel to the x-axis, the asymptotes will have the form:
y = ± (b/a) (x - h) + k
Plugging in the values from the standard form, we get:
y = ± (√(4225/9) / √(-4225/64)) (x + 8) - 8
y = ± (64/27) (x + 8) - 8
Simplifying and rounding to the nearest tenth, we get:
y = (2.4)x - 31.6 and y = (-2.4)x + 15.6 as the equations of the asymptotes.