Find the value of the term in the arithmetic sequence.

Mathematics

2 answers:

Vinvika [58]3 years ago

6 0

Answer:

its -18

Step-by-step explanation:

Lana71 [14]3 years ago

5 0

Answer:

i think the answer would be 18

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What expression is equivalent to 1/3(4-5x-1)?

Serhud [2]

1/3( 4 - 5x - 1)

= (1/3)(4 + - 5x + - 1)

= (1/3)( 4) + (1/3)(- 5x) + (1/3)( - 1)

= 4/3 + - 5x/3 + - 1/3

= - 5x/3 + 1 (Decimal: - 1.666667x + 1)

Hope that helps!!!!

3 0

3 years ago

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What is the slope of the line through (2, -1) and (-2, -3)?

prohojiy [21]

$\text{Given that,}\\\\\\(x_1,y_1) = (2,-1)~~\text{and}~~ (x_2,y_2) = (-2,-3)\\\\\\\text{Slope,}~ m =\dfrac{y_2-y_1}{x_2 -x_1} = \dfrac{-3+1}{-2-2} = \dfrac{-2}{-4} = \dfrac 12$

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3 years ago

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Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Marianna [84]

$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

7 0

3 years ago

Find the five arithmetic means between -18 and 36

Andreas93 [3]

Yeah it really be like that sometimes, you gotta try girl

5 0

3 years ago

Solve for f pleaseeeee

KonstantinChe [14]

Answer:

$\frac{3}{5}f = 4-2m\\ f= \frac{(4-2m) X 5}{3} \\f= \frac{20-10m}{3}$