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3 years ago

What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4

1 answer:

5 0

What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4

Need f(2):

f(2)=f(2-1)+4
f(2)=f(1)+4
f(2)=(3)+4=7

FIND f(3):
f(3)=f(3-1)+4
f(3)=f(2)+4
f(3)=(7)+4
f(3)=11

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You wish to test the following claim ( H a ) at a significance level of α = 0.01 . H o : μ = 60.8 H a : μ ≠ 60.8 You believe the

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3 years ago

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3 years ago

ABCD is a rhombus. The diagonals, AC and BD, intersect at the point M. The coordinates of M are (6, -11). The points A and C bot

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Answer:

The exact coordinates of the point where the line through B and D intersects the y-axis is $\left ( 0, \ -12\dfrac{5}{7} \right)$

Step-by-step explanation:

For the given question, we have the following parameters;

The shape of the quadrilateral ABCD in the question = A rhombus

The diagonals of the rhombus = AC and BD

The point of intersection of the diagonals, M = (6, -11)

The equation of the line points A and C lie = 2·y + 7·x = 20

A rhombus is an orthodiagonal quadrilateral (The diagonals of a rhombus are perpendicular)

Therefore;

AC ⊥ BD and where the slope of the line with points A and C is 'm', the slope of the line with points B and D is (-1/m)

Rewriting the equation, 2·y + 7·x = 20, in slope and intercept form, we have;

2·y + 7·x = 20

y = (- 7·x + 20)/2 = -7·x/2 + 10

∴ y = -7·x/2 + 10

The slope of the line 'y = -7·x/2 + 10', m = -7/2

∴ The slope of the line having the points B and D is -1/m = -1/(-7/2) = 2/7

The equation of the line having the points B and D in point and slope form, using point (6, -11) is given as follows;

y - (-11) = 2/7 × (x - 6)

y + 11 = 2·x/7 - 12/7

∴ y = 2·x/7 - 12/7 - 11 = 2·x/7 - 89/7 = 2·x/7 - $12\dfrac{5}{7}$

The equation of the line having the points B and D in point is therefore;

y = 2·x/7 - $12\dfrac{5}{7}$

The coordinates of the point where the line having the points B and D intersects the y-axis is the y-intercept, of the equation, y = 2·x/7 - $12\dfrac{5}{7}$ , which is $\left ( 0, \ -12\dfrac{5}{7} \right)$

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