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Pani-rosa [81]
3 years ago
8

If ∆JKL ≅ ∆PQR, which one of these is not a pair of corresponding parts? ​∠J and ∠P​ and ​∠L and ∠R​ and

Mathematics
1 answer:
Mandarinka [93]3 years ago
8 0
When trying to prove that two triangles are congruent there are three congruency rules SAS, ASA and SSS. From the looks of things though I would say it would be the third one because just from the way you named the triangles I could tell that the two angles could correspond.

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Identify the variable expression that is not a polynomial.
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B. 3\sqrt x -2

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Find the slope of the line. Write your answer in simplest form.
Savatey [412]

The slope of this line is: m = 3/4

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1. Approximate the given quantity using a Taylor polynomial with n3.
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Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

f(x) = x^{1/4}

The n-th order Taylor polynomial for function f with its center at a is:

p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}

As n = 3  So,

p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}

p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}

p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} }  (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} +  (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}

p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} }  (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} +  (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}

p_{3} (x) = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6  

                                                                                  (0.0000018522752) (x-81)³

p_{3} (x)  =  0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

                                                                                                       (x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

p_{3} (94)  =  0.0092592593 (94) - 0.000042866941 (94 - 81)² +          

                                                                 0.00000030871254 (94-81)³ + 2.25

         = 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +    

                                                                                                                       2.25

         = 0.87037 03742 - 0.000042866941 (169) +  

                                                                      0.00000030871254(2197) + 2.25

         = 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

p_{3} (94)  = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute \sqrt[4]{94} as 94^{1/4} using calculator

Exact value:

E_{a}(94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94)  = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94)  = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94)  = 0.0001

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3 years ago
Prove algebraically that the straight line with equation x=2y+5 is a tangent to the circle with equation x^2+y^2=5
zmey [24]

Differentiate both sides of the equation of the circle with respect to x, treating y=y(x) as a function of x:

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This gives the slope of any line tangent to the circle at the point (x,y).

Rewriting the given line in slope-intercept form tells us its slope is

x=2y+5\implies y=\dfrac12x-\dfrac52\implies\mathrm{slope}=\dfrac12

In order for this line to be tangent to the circle, it must intersect the circle at the point (x,y) such that

-\dfrac xy=\dfrac12\implies y=-2x

In the equation of the circle, we have

x^2+(-2x)^2=5x^2=5\implies x=\pm1\implies y=\mp2

If x=-1, then -1=2y+5\implies y=-3\neq2, so we omit this case.

If x=1, then 1=2y+5\implies y=-2, as expected. Therefore x=2y+5 is a tangent line to the circle x^2+y^2=5 at the point (1, -2).

7 0
3 years ago
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