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NNADVOKAT [17]
4 years ago
12

(3 points) In class we described the anti-malarial drug artemisinin (structure given below). What is the active functional group

on the drug that is most responsible for its potency?
Mathematics
1 answer:
Otrada [13]4 years ago
3 0

Answer: QINGHAOSU

Step-by-step explanation:Artemisinin is an antimalarial (Drug used to cure Malaria) drug whose active ingredient QINGHAOSU was isolated in the 1970s from the Plant called Artemisia annua by a Chinese Physician.

Artemisinin has been in use in ancient Chinese communities since the 4th century to cure Diseases. Artemisinin is potent in killing several forms of the Plasmodium specie and has been used to derive other antimalarial Drugs in use today like Artemeter and Artesunate.

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36z + 72 as a product of 2 factors
andreev551 [17]

Answer:

36(z + 2)

Step-by-step explanation:

First, find the GCF or greatest common factor. The GCF is the largest factor that all the numbers share. Both 72 and 36 are divisible by 36. 72 ÷ 36 = 2 and 36 ÷ 36 = 1. So, the factoring would look like 36(1z + 2) or 36 (z + 2)

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2 years ago
Translate 3 units to the right and then rotate 180 around the origin
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Answer to very first one: (-6,-4) (-10,4) (-5,3) (-4,-2)
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4 years ago
If (-2, y) lies on the graph of y = 3x, then y =?
algol13
Y=3x
you use the order pair to substitute therefore y= 3(-2)
y= -6 .
8 0
3 years ago
Read 2 more answers
Help please thank you <br> answer choices are: negatively skewed, symmetric, positively skewed
garri49 [273]

Answer:

the answer is symmetric i think

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
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