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3 years ago

The explicit formula for a sequence is

Mathematics

1 answer:

MAXImum [283]3 years ago

3 0

Step-by-step explanation:

$\because \: a_n = - 1 + 3(n - 1) \\ \\ \therefore \: a_{55} = - 1 + 3(55 - 1) \\ \\ \therefore \: a_{55} = - 1 + 3 \times 54 \\ \\ \therefore \: a_{55} = - 1 + 162\\ \\ \huge \blue { \boxed{\therefore \: a_{55} = 161}}\\ \\$

Hence, the 55th term of the sequence is 161.

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3 years ago

PLEASE HELP!! what is the equation for the parabolic function that goes through the points (-3, 67) (-1, 1) and has a stretch fa

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Answer:

$f(x)=9(x+\frac{1}{6})^2-\frac{21}{4}$

Step-by-step explanation:

So, we need to find the equation of a parabolic function that goes through the points (-3,67) and (-1,1) and has a stretch factor of 9.

In other words, we want to find a quadratic with a vertical stretch of 9 that goes through the points (-3,67) and (-1,1).

To do so, we first need to write some equations. Let's use the vertex form of the quadratic equation. The vertex form is:

$f(x)=a(x-h)^2+k$

Where a is the leading coefficient and (h,k) is the vertex.

Since a is the leading coefficient, it's also our stretch factor. Thus, let a equal 9.

Also, we have two points. We can interpret them as functions. In other words, (-3,67) means that f(-3) equals 67 and (-1,1) means that f(-1) equals 1. Write the two equations:

$f(x)=a(x-h)^2+k\\f(-3)=67=9((-3)-h)^2+k$

And:

$f(x)=a(x-h)^2+k\\f(-1)=1=9((-1)-h)^2+k$

Now, we essentially have a system of equations. Thus, to find the original equation, we just need to solve for the vertex. To do so, first isolate the k term in the second equation:

$1=9(-1-h)^2+k\\k=1-9(-1-h)^2$

Now, substitute this value to the first equation:

$67=9(-3-h)^2+k\\67=9(-3-h)^2+(1-9(-1-h)^2)$

And now, we just have to simplify.

First, from each of the square, factor out a negative 1:

$67=9((-1)(h+3))^2+(1-9(((-1)(h+1))^2)$

Power of a product property:

$67=9((-1)^2(h+3)^2)+(1-9((-1)^2(h+1)^2))$

The square of -1 is positive 1. Thus, we can ignore them:

$67=9(h+3)^2+(1-9(h+1)^2)$

Square them. Use the trinomial pattern:

$67=9(h^2+6h+9)+(1-9(h^2+2h+1))$

Distribute:

$67=(9h^2+54h+81)+(1-9h^2-18h-9)$

Combine like terms:

$67=(9h^2-9h^2)+(54h-18h)+(81+1-9)$

The first set cancels. Simplify the second and third:

$67=36h+73$

Subtract 73 from both sides. The right cancels:

$67-73=36h+73-73\\36h=-6$

Divide both sides by 36:

$(36h)/36=(-6)/36\\h=-1/6$

Therefore, h is -1/6.

Now, plug this back into the equation we isolated to solve for k:

$k=1-9(-1-h)^2$

First, remove the negative by simplifying:

$k=1-9((-1)(h+1))^2\\k=1-9((-1)^2(h+1)^2)\\k=1-9(h+1)^2$

Plug in -1/6 for h:

$k=1-9(-\frac{1}{6}+1)^2$

Add. Make 1 into 6/6:

$k=1-9(-\frac{1}{6}+\frac{6}{6})^2\\ k=1-9(\frac{5}{6})^2$

Square:

$k=1-9(\frac{25}{36})$

Multiply. Note that 36 is 9 times 4:

$k=1-9(\frac{25}{9\cdot4})\\ k=1-\frac{25}{4}$

Convert 1 into 4/4 and subtract:

$k=\frac{4}{4}-\frac{25}{4}\\ k=-\frac{21}{4}$

So, the vertex is (-1/6, -21/4).

Now, plug everything back into the very original equation with 9 as a:

$f(x)=a(x-h)^2+k\\f(x)=9(x+\frac{1}{6})^2-\frac{21}{4}$

And this is our answer :)

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