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rusak2 [61]
2 years ago
13

7. A vase 25 cm tall is positioned on a bench near a wall as shown. The shape of the vase follows the curve y = (x - 10)^2, wher

e y cm is the height of the vase and x cm is the distance of the vase from the wall.
a. How far is the base of the vase from the wall? b. What is the shortest distance from the top of the vase to the wall?
c. If the vase is moved so that the top just touches the wall, find the new distance from the wall to the base.
d. Find the new equation that follows the shape of the vase.​
Mathematics
1 answer:
Kitty [74]2 years ago
4 0

Answer:

Step-by-step explanation:

the base of the vase will be where the vase touches the x-axis, that is 10 cm, therefore, the base is 10 cm from the wall

:

b) 25 = x^2 -20x +100, we solve for x to find the closest distance since as we move up the vase the distance to the wall gets closer(assume the y-axis is the wall), then

x^2 -20x +75 = 0  (x-15) * (x-5) = 0

x = 15 and x = 5

we reject x = 15

the shortest distance from the top of the vase to the wall is 5 cm

:

c) this is a left shift of the equation y = (x-10)^2

from b) we know that the left shift is 5 cm

10 - 5 = 5 cm from the wall to the base

:

d) y = (x-10+5)^2

y = (x-5)^2    

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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 <em>0.002652</em> 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

<em>f(x)=cos(x) </em>

<em> </em>

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
Quick question
zaharov [31]

Answer:

False

Step-by-step explanation:

Let us evaluate the problem using the proper order of operation:

           4x − 2(5x − 1)

To solve this problem, we use the order of operation called PEMDAS;

P - Parentheses

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

Now, since operations inside the parentheses are of the utmost priority, let us solve this;

             4x − 2(5x − 1)

            = 4x - 10x +2

            = -6x + 2  or 2 - 6x

We see that the solution give is false.

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