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In-s [12.5K]
3 years ago
8

Find the nth term -1 8 19 32 47

Mathematics
1 answer:
koban [17]3 years ago
4 0

\bold{\text{Answer:}\quad \text{Recursive formula:}\ a_n=a_{-1}+2n+5}\\.\qquad \qquad \ \text{Explicit formula:}\ a_n=2n^2+3n-6

<u>Step-by-step explanation:</u>

-1 → 8 = +9

8 → 19 = +11

19 → 32 = +13

32 → 47 = + 15

a₁ = -1

d = 2n + 5

Recursive formula is: the previous term plus the difference (d)

                               \large\boxed{a_n=a_{n-1}+2n+5}

Explicit formula is the first term plus the product of d and n-1:

a_n=a_1+d(n-1)\\a_n=-1+(2n+5)(n-1)\\a_n=-1+2n^2-2n+5n-5\\\large\boxed{a_n=2n^2+3n-6}

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find the midpoint of the line segment whose endpoints are (-2, 5) and (4, -9). (2, -7) (1, -4) (1, -2)
Alex_Xolod [135]
M(x, y) = ((x1 + x2)/2, (y1 + y2)/2) = ((-2 + 4)/2, (5 - 9)/2) = (2/2, -4/2) = (1, -2)
7 0
3 years ago
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Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1
velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the i-th subinterval

\ell_i=\dfrac{i-1}5

r_i=\dfrac i5

where 1\le i\le5. The midpoint of each subinterval is

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}

Then the Riemann sum approximating the integral above is

2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5

\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)

\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}

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3 0
3 years ago
Given:
astraxan [27]

The answer is Vertical angles are equal.

5 0
3 years ago
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PLEASE HELP it should be easy for someone but I can’t do it
-Dominant- [34]

Answer:

Total crackers on the plate are 12

Step-by-step explanation:

Manuel ate crackers = \frac{1}{3}

His brother ate crackers = \frac{1}{4}

Crackers left on the plate = 5

We need to find how many crackers were there on the plate.

Let x be the total crackers on the plate

So, we can write the equation

x-\frac{1}{3}x-\frac{1}{4}x=5

Because 1/3 and 1/4 crackers are eaten and 5 are left so, we subtract 1/3x and 1/4x from x and equal it to 5

\frac{12x-4x-3x}{12}=5\\\frac{12x-7x}{12}=5\\\frac{5x}{12}=5\\x=\frac{5*12}{5}\\x=12

So, we get x = 12

Therefore, total crackers on the plate are 12

8 0
3 years ago
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