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Vinil7 [7]
3 years ago
14

On the first of three tests sally scored 75 points. on the third test her score was one point more than the second test her aver

age on the three test was 82 what were her scores on the second and third test
Mathematics
1 answer:
stepladder [879]3 years ago
3 0
If her second test was x and her third was x+1 (since it was 1 more than the second), we get 75+x+x+1=76+2x as our total test scores. Next, to get the average you add up the numbers and divide it by the amount of numbers you have. Since we have 3 tests and our sum is 76+2x, we divide it by 3 to get (76+2x)/3=average=82. Multiplying both sides by 3, we get 246=76+2x. Next, we subtract both sides by 76 to get 170=2x. Lastly, we divide both sides by 2 to get 85=x= the second test score. Her third test score is x+1=85+1=86
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3 years ago
A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m
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Answer:

The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

\Delta < 0\\b^{2}-4*a*c

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

\Delta < 0

-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\

4 0
3 years ago
What is 5/2x -5/2+7?
Elena L [17]
5x + 9 over 2

Explanation:
I know I’m right, but I hope this helped. Have a great day
4 0
3 years ago
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2x^2 / x^2 - y^2 - x^2 / xy - y ^ 2​
Juliette [100K]

Answer:

2 - 2y^2 - x/y

Step-by-step explanation:

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x^2/xy = x/y

therefore

2x^2/x^2 - y^2 - x^2/xy - y^2

= 2 - y^2 - x/y - y^2

= 2 - y^2 - y^2 - x/y

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6 0
3 years ago
3 cm
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Answer:

check my explanation

Step-by-step explanation:

yes it is c.

by adding the ones on screen

it is option C

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