Help please!!!!!!!!!!!!!

Mathematics

1 answer:

zhuklara [117]2 years ago

5 0

Answer= 13 _10= [23]_5
Please rewrite the same equation just put the 23 in the box

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15 Points!

Xelga [282]

$f(x)=\dfrac{x-5}{x^2-1}\\\\\text{the horizontal asymptotes}\\\\x^2-1=0\to x^2=1\to x=\pm\sqrt1\to x=-1\ \wedge\ x=1\\\\\text{the vertical asymptotes}\\\\y=\lim\limits_{x\to\pm\infty}\dfrac{x-5}{x^2-1}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x}{x^2}+\frac{5}{x^2}}{\frac{x^2}{x^2}-\frac{1}{x^2}}=\dfrac{0+0}{1-0}=0$

Answer: <span>B) x = 1, x = -1, y = 0</span>

8 0

3 years ago

Compute i^1+i^2+i^3....i^99+i^100

fgiga [73]

Good morning ☕️

Answer:

Step-by-step explanation:

Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰

S = i¹ + i² + i³ + . . . + i⁹⁹ + i¹⁰⁰

S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀

then, S is the sum of 100 consecutive terms of a geometric sequence (an)

where the first term a1 = i¹ = i and the common ratio = i