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Angelina_Jolie [31]
3 years ago
6

Which best describes this triangle? A. All sides are the same length; all angles are acute. B. Two sides are the same length; on

e angle is obtuse. C. Two sides are the same length; all angles are acute. D. All sides are different lengths; all angles are acute.
Mathematics
2 answers:
Ilya [14]3 years ago
7 0
None are correct because not all angles of the triangle are acute because one side of it is a right angle and not one of the angles or obtuse. <span />
DanielleElmas [232]3 years ago
3 0
C. Two sides are the same length; all angles are acute
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What is the markup percent on a diamond for which the markup is $870 and the selling price is $2870?
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Answer:

30.31%

Step-by-step explanation:

Mark up on selling price is given by markup*100%/selling price

In this case, markup is given as $870 while the selling price is $2870 hence the percentage of narkup to selling price will be given by \frac {870\times 100}{2870}=30.313588850174\approx 30.31

Therefore, the percentage of markup to selling price is approximately 30.31%

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Please read the attached Newsela article, "Pets can make people feel happier and healthier" and complete the attached Double Ent
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Analyzing a Graph In Exercise, analyze and sketch the graph of the function. Lable any relative extrema, points of inflection, a
Sladkaya [172]

Answer:

y=(\ln{x})^2

point of extremity: (1,0)

vertical asymptote: along the y-axis (x = 0)

point of inflection: (e,1)

Solution:

Although all these points can be directly observed from the graph below, but these are the analytical solutions if you're curious!

1) Extreme point can be found by differentiating 'y' once and equating to zero. solving for x:

\dfrac{dy}{dx}=\dfrac{dy}{dx}((\ln{x})^2)

\dfrac{dy}{dx}=2\ln{x}\left(\dfrac{1}{x}\right)

substitute dy/dx = 0, and solve for x

0=2\ln{x}\left(\dfrac{1}{x}\right)

0=2\ln{x}

x=e^0

x=1

use this value of x back in y, to find the y-coordinate of the extreme point

y=(\ln{1})^2

y=0

The extreme point = (1,0)

2) Differentiate y twice to find the inflection point.

\dfrac{dy}{dx}=2\ln{x}\left(\dfrac{1}{x}\right)

\dfrac{d^2y}{dx^2}=2\ln{x}\left(-\dfrac{1}{x^2}\right)+\left(\dfrac{1}{x}\right)\left(\dfrac{2}{x}\right)

\dfrac{d^2y}{dx^2}=\dfrac{2}{x^2}\left(-\ln{x}+1}\right)

substitute d2y/dx2 = 0, and solve for x

0=\dfrac{2}{x^2}\left(-\ln{x}+1}\right)

0=-\ln{x}+1

\ln{x}=1

x = e

use this value of x back in y, to find the y-coordinate of the inflection point

y=(\ln{e})^2

y=1

The extreme point = (e,1)

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