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3 years ago

Find the value of x round to the nearest tenth

Mathematics

2 answers:

pychu [463]3 years ago

7 0

Answer:

23.0 =x

Step-by-step explanation:

Assuming this is a right triangle

cos theta = adj/ hypotenuse

cos 23 = x/25

Multiply each side by 25

25 cos 23 = x

23.01262134 =x

To the nearest tenth

23.0 =x

Elina [12.6K]3 years ago

6 0

23.0 i’m sure that’s the answer

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Find the median of these numbers 5,9,13,7,5,7

Troyanec [42]

Answer:

The mean is 7

Step-by-step explanation:

5,5,7,7,9,13

9-7 is going to get 7

3 0

3 years ago

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

3 years ago

Question 1 (Multiple Choice Worth 1 points)

Oksi-84 [34.3K]

Sorry I don't know but i bet more can exolain.

4 0

3 years ago

In a flower garden, there are 4 tulips for every 9 daisies. If there are 24 tulips, how many daisies are there? Please help i wi

miss Akunina [59]

54 daisies is the answer

Hope you have a good day!

4 0

3 years ago

Suppose a regular n-gon is inscribed in a circle of radius r. Diagrams are shown for n=6, n=8, and n = 12. n = 6 n=8 n = 12 a a

m_a_m_a [10]

Answer:If an N-gon (polygon with N sides) has perimeter P, then each of the

N sides has length P/N. If we connect two adjacent vertices to the

center, the angle between these two lines is 360/N degrees, or 2*pi/N

radians. (Do you understand radian measure well enough to follow me

this far?)

The two lines I just drew, plus the side of the polygon between them,

form an isosceles triangle. Adding the altitude of the isosceles

triangle makes two right triangles, and we can use one of them to

derive the equation

sin(theta/2) = s/(2R)

where theta is the apex angle (which I said is 2*pi/N radians), R is

the length of the lines to the center (the radius of the circumscribed

circle), and s is the length of the side (which I said is P/N).

Putting those values into the equation, we have

sin(pi/N) = P/(2NR)

so that

P = 2NR sin(pi/N)

gives the perimeter of the N-gon with circumradius R.

Can we see a connection between this formula and the perimeter of a

circle? The perimeter of the circumcircle is 2*pi*R. As we increase

N, the perimeter of the polygon should get closer and closer to this

value. Comparing the two, we see

2NR*sin(pi/N) approaches 2*pi*R

N*sin(pi/N) approaches pi

You can check this out with a calculator, using big numbers for N such

as your teacher's N=1000. If you calculate the sine of an angle in

degrees rather than radians, the formula will look like

N*sin(180/N) --> pi

Step-by-step explanation:

6 0

3 years ago