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2 years ago

6, -18, 54, ... what is the equation for the sequence give the 14th term

Mathematics

1 answer:

Rom4ik [11]2 years ago

7 0

Answer:

-9,565,938

Step-by-step explanation:

The first term of this geometric sequence is 6. Its common ratio is -3. The n-th term is given by ...

an = (6)(-3)^(n-1)

Then the 14th term is ...

a14 = 6·(-3)^(14-1)

a14 = -9,565,938

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mixer [17]

When ever you have percentages, it should be helpful to bear in mind you can express them as multipliers. In this case, it will be helpful.
So, if we let:
a = test score
b = target score
then, using the information given:
a = 1.1b + 1
a = 1.15b - 3
and we get simultaneous equations.
'1.1' and '1.15' are the multipliers that I got using the percentages. Multiplying a value by 1.1 is the equivalent of increasing the value by 10%. If you multiplied it by 0.1 (which is the same as dividing by 10), you would get just 10% of the value.
Back to the simultaneous equations, we can just solve them now:
There are a number of ways to do this but I will use my preferred method:
Rearrange to express in terms of b:
a = 1.1b + 1
then b = (a - 1)/1.1
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then b = (a + 3)/1.15
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(a - 1)/1.1 = (a + 3)/1.15
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2 years ago

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2 years ago

Indicate the equation of the given line in standard form. Show all your work for full credit. the line containing the median of

alukav5142 [94]

Answer:

* The equation of the median of the trapezoid is 10x + 6y = 39

Step-by-step explanation:

* Lets explain how to solve the problem

- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

- The standard form of the linear equation is Ax + BC = C, where

A , B , C are integers and A , B ≠ 0

- The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

- It has two properties:

# It is parallel to both bases

# Its length equals half the sum of the base lengths

* Lets solve the problem

- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)

- Lets find the slope of the 4 sides two find which of them are the

parallel bases and which of them are the non-parallel bases

# The side RS

∵ $m_{RS}=\frac{8-5}{1 - (-1)}=\frac{3}{2}$

# The side ST

∵ $m_{ST}=\frac{-2-8}{7-1}=\frac{-10}{6}=\frac{-5}{3}$

# The side TU

∵ $m_{TU}=\frac{0-(-2)}{2-7}=\frac{2}{-5}=\frac{-2}{5}$

# The side UR

∵ $m_{UR}=\frac{5-0}{-1-2}=\frac{5}{-3}=\frac{-5}{3}$

∵ The slope of ST = the slop UR

∴ ST// UR

∴ The parallel bases are ST and UR

∴ The nonparallel sides are RS and TU

- Lets find the midpoint of RS and TU to find the equation of the

median of the trapezoid

∵ The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

∵ The midpoint of RS = $(\frac{-1+1}{2},\frac{5+8}{2})=(0,\frac{13}{2})$

∵ The median is parallel to both bases

∴ The slope of the median equal the slopes of the parallel bases = -5/3

∵ The form of the equation of a line is y = mx + c

∴ The equation of the median is y = -5/3 x + c

- To find c substitute x , y in the equation by the coordinates of the

midpoint of RS

∵ The mid point of Rs is (0 , 13/2)

∴ 13/2 = -5/3 (0) + c

∴ 13/2 = c

∴ The equation of the median is y = -5/3 x + 13/2

- Multiply the two sides by 6 to cancel the denominator

∴ The equation of the median is 6y = -10x + 39

- Add 10x to both sides

∴ The equation of the median is 10x + 6y = 39

* The equation of the median of the trapezoid is 10x + 6y = 39

7 0

3 years ago