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Andrew [12]
3 years ago
10

Solve each system of linear equations albebraically

Mathematics
1 answer:
ANEK [815]3 years ago
5 0

Step-by-step explanation:

1.\\\left\{\begin{array}{ccc}y=3x&(1)\\2y=6x&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\2(3x)=6x\\6x=6x\qquad\text{subtract}\ 6x\ \text{from both sides}\\6x-6x=6x-6x\\0=0\qquad\text{TRUE}\\\\Answer:\ \text{Infinitely many solutions}\\\\\left\{\begin{array}{ccc}y=3x\\x\in\mathbb{R}\end{array}\right

2.\\\left\{\begin{array}{ccc}y=2x+5&(1)\\y-2x=1&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\(2x+5)-2x=1\\2x+5-2x=1\qquad\text{combine like terms}\\(2x-2x)+5=1\\5=-1\qquad\text{FALSE}\\\\Answer:\ \text{No solution.}

3.\\\left\{\begin{array}{ccc}3x-2y=9\\-6x+4y=1&\text{divide both sides by 2}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}3x-2y=9\\-3x+2y=0.5\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad0=9.5\qquad\text{FALSE}\\\\Answer:\ \text{No solution.}

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Compute i^1+i^2+i^3....i^99+i^100
fgiga [73]

Good morning ☕️

Answer:

<h3>i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰ = 0</h3>

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

FORMULA:______________________

S=(term1)*\frac{1-(common.ratio)^{number.of.terms}}{1-(common.ratio)}

_______________________________

then

S=i*\frac{1-i^{100} }{1-i}

or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1  (we know that i⁴ = 1)

Hence

S = 0

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3 years ago
PLEASE HELP FAST!<br><br><br> what is the length of segment AB? <br> 10<br> 12<br> 13<br> 15
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Answer:

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Step-by-step explanation:

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Using the pythagorean theorem, a^2 + b^2 = c^2, we get 25 + 144 = 169 or 13^2.

Therefore the answer is 13.

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