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UkoKoshka [18]
2 years ago
15

Find the sum of the first five terms in the series

Mathematics
1 answer:
iragen [17]2 years ago
4 0
122.

The numbers are multiplied by -3 every time, so the fifth term would be 162, then you add the numbers all together to make 122.
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How to classify 8x4 + 7x3 + 5x2 + 8 by number of terms.
Arisa [49]

Answer:

This is a polynomial.

Step-by-step explanation:

A monomial is a constant, a variable or the product of constants and variables.  

A binomial is the sum or difference of two monomials.

A trinomial is the sum or difference of three monomials.

Anything more than three terms makes it a polynomial.

This has 4 terms; thus it is a polynomial.

4 0
3 years ago
Read 2 more answers
Y = (x) =(1/8)^x<br> Find f(x) when X =<br> (1/3)<br> Round your answer to the nearest thousandth.
vlada-n [284]

Answer:

1/2.

Step-by-step explanation:

f(x) = (1/8)^x

when x = 1/3

f(x)  =  (1/8)^1/3

f(x) = ∛(1/8)

f(x) = 1 / ∛8

f(x) = 1/2.

6 0
3 years ago
How to find the coordinates?
zhenek [66]
The lower number of the fraction is the X value and since it's negative u go 3 to the right(if u loon at point T)and since it's a negative 4 on the top you go 4 down and if you would reverse it you would notice that 0|4 is 1 y value below the straight line so the coordinates are (3|5)
6 0
3 years ago
What’s the median of this data set ?
earnstyle [38]

Answer:

56

Step-by-step explanation:

The median is the middle number of a sorted set from least to greatest. After sorting the set 61, 48, 52, 65, 56, 77, 54 the set is 48, 52, 54, 56, 61, 65, 77.

The middle number in this set is 56 so the median is 56

3 0
2 years ago
Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)
Triss [41]

g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}

The surface element is

\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

and the integral is

\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)

###

To compute the last integral, you can integrate by parts:

u=r\implies\mathrm du=\mathrm dr

\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}

\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr

For this integral, consider a substitution of

r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds

\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds

\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds

=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds

and the result above follows.

4 0
3 years ago
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