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3 years ago

Write the following two sequences using summation notation:

Mathematics

1 answer:

o-na [289]3 years ago

7 0

Answer:

Step-by-step explanation:

Sequence 1:

This is a finite sequence ending with 40

5-10+15-20+...40

= 5(1-2+3-4+5-6+7-8) (by taking 5 as common factor)

This is an alternating series with even numbers in the series having negative sign

Hence can be written as

$\Sigma _{1} ^{8} 5[(-1)^{n-1} (n)$

This gives the I sequence in the summation notation. Since there are only 8 terms n can take values as 1 to 8, with odd terms positive.

---------------------------------

Sequence 2:

This is the same as the previous sequence but with end point as infinite

This is an infinite series. Hence n can vary from 1 to infinity.

Thus sum would be

$\Sigma _{1} ^{\infty } 5[(-1)^{n-1} (n)$

(Note that only the ending number is difference for both.)

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Answer:

Step-by-step explanation:

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36-9=27

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x=5.2

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3 years ago

Walter walked 15.5 blocks from his house to work. It took him 35 minutes. What is his rate in blocks per hour?

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3 years ago

20 points pls help quick

masha68 [24]

Answer:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

Step-by-step explanation:

We are given the two functions:

$\displaystyle f(x) = 3x^2 + mx +5 \text{ and } g(x) = nx^2 - 4x -2$

And that:

$\displaystyle h(x) = f(x)\cdot g(x)$

With the given conditions that (1, -40) and (-1, 24) satisfy the new function, we want to determine functions f and g.

First, find h:

$\displaystyle \begin{aligned} h(x) & = f(x)\cdot g(x) \\ \\ & = (3x^2 + mx +5)(nx^2 - 4x -2) \end{aligned}$

Because (1, -40) and (-1, 24) are points on the graph of h, we have that h(-1) = 40 and h(-1) = 24. In other words:

$\displaystyle \begin{aligned} h(1) = -40 & = (3(1)^2 + m(1) +5)(n(1)^2 - 4(1) -2) \\ \\ & = (3 + m +5)(n-4 -2) \\ \\ & = (m+8)(n-6) \\ \\ -40 &= mn-6m+8n-48 \end{aligned}$

And:

$\displaystyle \begin{aligned} h(-1) = 24 & = (3(-1)^2 + m(-1) +5)(n(-1)^2 -4(-1) -2) \\ \\ & = (3 - m +5)(n + 4 -2) \\ \\ & = (-m+8)(n+2) \\ \\ 24 & = -mn -2m + 8n +16 \end{aligned}$

Solve the system of equations. Adding the two equations together yield:

$\displaystyle -16 = -8m+16n - 32$

Solve for either m or n:

$\displaystyle \begin{aligned} -16 & = -8m + 16n - 32 \\ \\ 16 & = -8m + 16n \\ \\ 8m & = 16n - 16 \\ \\ m & = 2n -2\end{aligned}$

Substitute this into one of the two equations above and solve:

$\displaystyle \begin{aligned} -40 & = mn - 6m + 8n - 48 \\ \\ 0 & = (2n-2)n -6 (2n-2) + 8n -8 \\ \\ &= (2n^2 - 2n) + (-12n + 12) +8 n - 8 \\ \\ & = 2n^2 -6n + 4 \\ \\ & = n^2 - 3n + 2 \\ \\ &= (n-2)(n-1) \\ \\ & \end{aligned}$

Therefore:

$\displaystyle n = 2 \text{ or } n = 1$

Solve for m:

$\displaystyle \begin{aligned}m &= 2n-2 & \text{ or } m & = 2n-2 \\ \\ & = 2(2) - 1 &\text{ or } & =2(1) -2 \\ \\ &= 2 &\text{ or } & = 0 \end{aligned}$

Hence, the values of n and m are either: 2 and 2, respectively; or 1 and 0, respectively.

In conclusion, functions f and g are:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

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2 years ago

For the function f(x) = 2 - 3x, find f(-4). Options are in the picture.

Kitty [74]

You just have to plug in -4 into
$f(x) = 2 - 3x$
This implies that,
$f( - 4) = 2 - 3( - 4)$

$f( - 4) = 2 + 3(4)$
$f( - 4) = 2 + 12$
$f( - 4) = 14$

4 0

3 years ago

Is AABC-ADEP? If so, identify the similarity postulate or theorem that

Mariana [72]

Answer:

D. cannot be determined

Step-by-step explanation:

that is kind of a trick question.

right based on the given information none of the 3 methods can be applied.

we don't have all 3 sides for SSS

we don't have 2 angles for AA

and we don't have 2 sides with their included angle for SAS.

but it can still be determined (they ARE similar), as this is a special case of SSA, where it is clear that the corresponding congruent angles are located at the same position relatively to the corresponding similar lines.

you can discuss this with your teacher, if you are interested.

AND - we can easily determine any of the other missing pieces (side and angles). these missing pieces then will correspond exactly to their counterparts in the other triangle (the 2 angles in ABC are equal to the 2 angles in DEF, the missing sides file the same scaling factor as the other known sides). and then any of the 3 methods can be applied.

so, this is actually tricky ...

3 0

3 years ago