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katen-ka-za [31]
3 years ago
11

Check the screenshot below ↓

Mathematics
2 answers:
azamat3 years ago
5 0
Y²-12y + 35 = 0

y² - 12y = -35

y²-12y + 36 = -35 + 36

(y-6)² = 1

y-6 = ± 1

y-6 = -1

y = -1 + 6

y = 5


y-6 = 1

y = 1+6

y = 7


y = 5 and y = 7
mrs_skeptik [129]3 years ago
3 0

Answer: y = 7, 5

Step-by-step explanation:

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What is the difference between a factor of a number and a multiple of a number? use the number 25 has an example
Ne4ueva [31]

Answer:

  • factor: a submultiple; an integer value that gives an integer quotient when the number is divided by it.
  • multiple: the product of the number and another integer

Step-by-step explanation:

<h3>Factors</h3>

A set of factors of a number (N) is a set of integers {f1, f2, f3, ...} whose product is the number:

  f1 × f2 × f3 × ... = N

Often the term "factors" is used to mean "prime factors," the set of prime numbers whose product is N.

When N = 25, the prime factors are {5, 5}. That is ...

  5×5 = 25

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<h3>Divisors</h3>

A "divisor" of a number is a sub-multiple of N. That is, the quotient N/k is an integer for some divisor k of N. Usually, we're interested in integer divisors. All prime factors will be integer divisors of N. The term "factor" is often used when the term "divisor" is meant.

Divisors of N = 25 include 1, 5, and 25. There will be an odd number of integer divisors (as here) if the number N is a perfect square.

__

<h3>Multiples</h3>

For an integer k, the value k×N is called a <em>multiple</em> of N, often, the <em>k-th multiple</em> of N.

Multiples of 25 include 25, 50, 75, 100, 125, and any other decimal number ending in 25, 50, 75, or 00.

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<em>Another example</em>

When N=30, the prime factors are {2, 3, 5}. The divisors are {1, 2, 3, 5, 6, 10, 15, 30}. (Note there are an even number of divisors.) Multiples of 30 include 30, 60, 90, 120, ....

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The answer for your specific question is G.
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What is the solution to this system of linear equations?2x + y = 13x – y = –6(–1, 3)(1, –1)(2, 3)(5, 0)
worty [1.4K]

\underline{+\left\{\begin{array}{ccc}2x+y=1\\3x-y=-6\end{array}\right}\ \ \ \ |\text{add both sides of equations}\\.\ \ \ \ \ 5x=-5\ \ \ \ |:5\\.\ \ \ \ \ \ x=-1\\\\\text{put the value of x to the first equation}\\\\2(-1)+y=1\\-2+y=1\ \ \ |+2\\y=3\\\\\text{Answer:}\ (-1,\ 3)

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