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lukranit [14]
3 years ago
7

a(t)=(t−k)(t−3)(t−6)(t+3) is a polynomial function of tt, where kk is a constant. Given that a(2)=0a(2)=0, what is the absolute

value of the product of the zeros of aa? Answer:
Mathematics
1 answer:
Crank3 years ago
3 0

Since a(2)=0, we know that t-2 must be a factor of a(t), so k=2. Then the zeros of a(t) are t=2,3,6,-3, and their product is -108, whose absolute value is 108.

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Answer:

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Step-by-step explanation:

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What is the mean of 4,6,7,7,11,12,15,18,20
ExtremeBDS [4]

Answer:

<h2>11.1111111111 or 1.1 <-(rounded)</h2>

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I hope this helped

4 0
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Daryl invested $2,200 for 3 years. He received interest of $264. What was the interest rate?
Klio2033 [76]

Answer:

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Step-by-step explanation:

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Solving multi step equations
hichkok12 [17]

The no-brain way to do it is to

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  2. divide by the coefficient of the variable.
  3. add the opposite of the constant.

Of course, at some point, you need to simplify the equation so you have something like

... ax + b = 0 . . . . . . where <em>a</em> and <em>b</em> are some constants that may be positive or negative

17) Subtract the right side.

... 10(x +3) -(-9x -4) -(x -5 +3) = 0

... 10x + 30 +9x +4 -x +5 -3 = 0 . . . . . eliminate parentheses

... x(10+9-1) +(30+4+5-3) = 0 . . . . . . . collect terms

... 18x +36 = 0 . . . . . . . . . . . . . . . . . . . simplified

... x + 2 = 0 . . . . . . . . . . . . . . . . . . . . . .divide by 18, the coefficient of x

... x = -2 . . . . . . . . . . . . . . . . . . . . . . . . add the opposite of the constant

19) Add the opposite of the left side.

... 0 = -9(1 +7x) +12(x -12)

... 0 = -9 -63x +12x -144 . . . . . . eliminate parenthses

... 0 = -51x -153 . . . . . . . . . . . . . simplify

... 0 = x +3 . . . . . . . . . . . . . . . . . divide by -51, the coefficient of x

... -3 = x . . . . . . . . . . . . . . . . . . . add the opposite of the constant

_____

If you examine the variable's coefficients you can make a choice of side to subtract that results in a positive coefficient of the variable.

This method puts variable and constant together until the end. The approach usually taught is to separate the variable terms and constant terms. (The number of steps required is the same either way.)

The reason this is "no brain" is that it always works and requires no judgment as to what you add or subtract from where. Applying a little judgment as described above can make it so you're mostly working with positive numbers, but the method works whether the numbers are positive or negative.

6 0
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If x = y and y=z which statement is true?
dybincka [34]
The x=y
Is this the the test of the question?
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