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inna [77]
3 years ago
12

Question 10 of 10

Mathematics
1 answer:
Reil [10]3 years ago
4 0

please help me out please help me out please help me out

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Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...​
rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

x_n = a n^4 + b n^3 + c n^2 + d n + e

From the given known values of the sequence, we have

\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}

Solving the system yields coefficients

a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4

so that the n-th term in the sequence might be

\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}

Then the next few terms in the sequence could very well be

\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of \{x_n\} eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of \{x_n\} by \Delta^{k}\{x_n\}. Then

• 1st-order differences:

\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}

• 2nd-order differences:

\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}

• 3rd-order differences:

\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}

• 4th-order differences:

\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}

From here I made the assumption that \Delta^4\{x_n\} is the constant sequence {15, 15, 15, …}. This implies \Delta^3\{x_n\} forms an arithmetic/linear sequence, which implies \Delta^2\{x_n\} forms a quadratic sequence, and so on up \{x_n\} forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

5 0
2 years ago
Solve by substitution 9x+y=21, -x-3=y
VashaNatasha [74]

Answer:

point form: (3,-6)

equation form: x=3, y=-6

Step-by-step explanation:

4 0
2 years ago
The volume of a rectangular prism is 50 5/8 cubic inches.
Temka [501]

mcyt mcyt mcyt mcyt mcyt

6 0
2 years ago
Evaluate the function f(x) at the given numbers (correct to six decimal places). F(x) = x2 − 5x x2 − x − 20 , x = 5.5, 5.1, 5.05
exis [7]

we are given

f(x)=\frac{x^2-5x}{x^2-x-20}

Firstly, we will simplify it

f(x)=\frac{x(x-5)}{(x-5)(x+4)}

f(x)=\frac{x}{(x+4)}

At x=5.5:

we can plug x=5.5

f(5.5)=\frac{5.5}{(5.5+4)}

f(5.5)=0.578947

At x=5.1:

we can plug x=5.1

f(5.1)=\frac{5.1}{(5.1+4)}

f(5.1)=0.56043956

At x=5.05:

we can plug x=5.05

f(5.05)=\frac{5.05}{(5.05+4)}

f(5.05)=0.558011

At x=5.01:

we can plug x=5.01

f(5.01)=\frac{5.01}{(5.01+4)}

f(5.01)=0.5560488

At x=5.005:

we can plug x=5.005

f(5.005)=\frac{5.005}{(5.005+4)}

f(5.005)=0.5558023

At x=5.001:

we can plug x=5.001

f(5.001)=\frac{5.001}{(5.001+4)}

f(5.001)=0.5556049

At x=4.9:

we can plug x=4.9

f(4.9)=\frac{4.9}{(4.9+4)}

f(4.9)=0.5505617

At x=4.95:

we can plug x=4.95

f(4.95)=\frac{4.95}{(4.95+4)}

f(4.95)=0.5530726

At x=4.99:

we can plug x=4.99

f(4.99)=\frac{4.99}{(4.99+4)}

f(4.99)=0.55506117

At x=4.995:

we can plug x=4.995

f(4.995)=\frac{4.995}{(4.995+4)}

f(4.995)=0.55503085

At x=4.999:

we can plug x=4.999

f(4.999)=\frac{4.999}{(4.999+4)}

f(4.999)=0.55550616



7 0
3 years ago
6. A sector of a circle is a region bound by an arc and the two radii that share the arc's endpoints. Suppose you have a dartboa
Aliun [14]

Given the dartboard of diameter 20in, divided into 20 congruent sectors,

  • The central angle is 18^\circ
  • The fraction of a circle taken up by one sector is \frac{1}{20}
  • The area of one sector is 15.7in^2 to the nearest tenth

The area of a circle is given by the formula

A=\pi r^2

A sector of a circle is a fraction of a circle. The fraction is given by \frac{\theta}{360^\circ}. Where \theta is the angle subtended by the sector at the center of the circle.

The formula for computing the area of a sector, given the angle at the center is

A_s=\dfrac{\theta}{360^\circ}\times \pi r^2

<h3>Given information</h3>

We given a circle (the dartboard) with diameter of 20in, divided into 20 equal(or, congruent) sectors

<h3>Part I: Finding the central angle</h3>

To find the central angle, divide 360^\circ by the number of sectors. Let \alpha denote the central angle, then

\alpha=\dfrac{360^\circ}{20}\\\\\alpha=18^\circ

<h3>Part II: Find the fraction of the circle that one sector takes</h3>

The fraction of the circle that one sector takes up is found by dividing the angle a sector takes up by 360^\circ. The angle has already been computed in Part I (the central angle, \alpha). The fraction is

f=\dfrac{\alpha}{360^\circ}\\\\f=\dfrac{18^\circ}{360^\circ}=\dfrac{1}{20}

<h3>Part III: Find the area of one sector to the nearest tenth</h3>

The area of one sector can be gotten by multiplying the fraction gotten from Part II, with the area formula. That is

A_s=f\times \pi r^2\\=\dfrac{1}{20}\times3.14\times\left(\dfrac{20}{2}\right)^2\\\\=\dfrac{1}{20}\times3.14\times10^2=15.7in^2

Learn more about sectors of a circle brainly.com/question/3432053

8 0
2 years ago
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