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3 years ago

The HA theorem is a special case of the ????

Mathematics

2 answers:

Whitepunk [10]3 years ago

8 0

The HA Theorem is a special case of the AAS postulate.

dybincka [34]3 years ago

4 0

Answer:

AAS theorem

Step-by-step explanation:

apex

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Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

2 years ago

3. Kanyla makes and

Triss [41]

Answer:

$23

Step-by-step explanation:

1 phone case = $12

4 phone cases = 12 * 4 = $48

Table = $25

4 phone cases - Table = 48 - 25 = $23

4 0

3 years ago

Read 2 more answers

A circle is centered at N(-6, -2). The point E(-1,1) is on the circle.

galina1969 [7]

Answer:

Outside the circle

Step-by-step explanation:

Let's first write the equation of this circle: $(x-h)^2+(y-k)^2=r^2$ , where (h, k) is the center and r is the radius. Here, the center is (-6, -2). We need to find the radius, which will just be the distance from N to E:

NE = $\sqrt{(-6-(-1))^2+(-2-1)^2} =\sqrt{25+9} =\sqrt{34}$

The radius is √34, which means that r² = 34. So, our equation is:

(x + 6)² + (y + 2)² = 34

Plug in -10 for x and -7 for y:

(x + 6)² + (y + 2)² = 34

x² + 12x + 36 + y² + 4y + 4 = 34

x² + 12x + y² + 4y + 40 = 34

x² + 12x + y² + 4y + 6 = 0

(-10)² + 12 * (-10) + (-7)² + 4 * (-7) + 6 = 7

Since 7 > 0, we know that H lies outside the circle.

6 0

3 years ago

Read 2 more answers

Sketch the graph of each line y=-9x-5

ycow [4]

I can’t answer it without a picture of the graph

5 0

3 years ago

Find the product of (x-7)^2 and explain how it demonstrates the closure property of multiplication

iogann1982 [59]

A. x²-14x+49; is a polynomial

Step-by-step explanation:

(x-7)² can be written as (x-7)(x-7)

Expanding the expression

x(x-7)-7(x-7)

x²-7x-7x+49

x²-14x+49 ⇒⇒A quadratic function, which is a polynomial of degree 2

This function demonstrates the closer property of multiplication in that the change in order of multiplication does not change the product. This is called commutative property.

(x-7)(x-7)

-7(x-7)+x(x-7)

-7x+49+x²-7x

x²-14x+49

Learn More

Polynomials :brainly.com/question/9601478

Keywords : product, closure property of multiplication,

#LearnwithBrainly

4 0

2 years ago