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gavmur [86]
4 years ago
13

Write a quadratic equations given the roots 5/2 and 9

Mathematics
1 answer:
sukhopar [10]4 years ago
7 0

Answer:

(x-5/2)(x-9).

x^2-5/2x-5/2x+45/2.

Multiply the number by 2.

which is 2(x^2-5/2x-5/2x+45/2).

2x^2-5x-5x+45.

2x^2-10x+45.

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Is 0 a positive number?
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Yes 0 is a positive number ! :)


4 0
4 years ago
Read 2 more answers
You began a shopping spree with $70 in your checking account. You bought a pair of headphones for $19.95. Then, you bought five
nikklg [1K]

Answer:

$102.55

Step-by-step explanation:

70 - 19.95 - (10.50 X 5) + 105 = 102.55

5 0
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
____ [38]

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
The first ice cream cone was made at the World’s Fair in St. Louis in 1904 when an ice cream seller ran out of cups. Suppose a s
irina1246 [14]
Volume of cone: pi x r^2 x H/3
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 volume of ice cream:  4/3 x pi x R^3
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volume of cone is larger than volume of ice cream so no it will not overflow


8 0
3 years ago
Answer ASAP please!<br> ….
VladimirAG [237]
For triangle a use Pythagorean theorem to find the hypotenuse.
A^2+b^2=c^2
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For b) 18^2+B^2=23^2
324+B^2=529
B^2=205
B=14.32 or square root of 205
8 0
3 years ago
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