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2 years ago

Provide the missing reasons for the proof. Given: AB = CB EB DB Prove: ABC ~ EBD

Mathematics

1 answer:

Umnica [9.8K]2 years ago

3 0

Answer:

Number 1 is Given.

Number 2 is the corresponding parts theorem

AB and CB are similar

also EB and DB

Number 3 is SSA similarly theorem

Step-by-step explanation:

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Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

Evaluate the function at the point: $f\left(2\right)=\frac{7}{3}$

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}$

Apply the Taylor series definition:

$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

8 0

3 years ago

An arrow is shot vertically upward from a platform 16ft high at a rate of 190ft/sec. When will the arrow hit the ground?

natta225 [31]

Answer:

time required for arrow to reach ground is 10.8 sec

when t = 10.8 seconds to the nearest tenth

Step-by-step explanation:

Given values are

Velocity = 190 ft/sec height = 16 ft

Given formula is

h=−16t²+vt+h0

adding the values, we get

h(t)= -16t²+190t+16

so we have to find when the hit he ground, so at ground the height will be 0

0= -16t²+190t+16

-16t²² + 190t + 16= 0

using the quadratic formula

x = (-b +(b2 - 4ac)1/2)/2a

values

a = -16, b = 190, c = 16

x= -190 + ((-190)²- 4 (-16)(16) ½ / 2(-16)

x= -190 + ((-190) ²+1024) ½ / 2(-16)

x= -190+ (-190²+ 32²) ½ /-32

taking under root

x= -190 + ( -190+32)/ -32

x= 190 + (-158)/-32

we have 2 options,

x= 190 + (-158)/-32 x= 190 - (-158)/-32

x= 190 -158)/-32 x= 190 +158)/-32

x= -1 x= 10.875

or t = 10.875

so the negative root has no practical significance

and we have t = 10.8 seconds to the nearest tenth

4 0

3 years ago

Ryan bought a digital camera with a list price of $150 from an online store offering a 5% discount. He needs to pay $5.50 for sh

zhannawk [14.2K]

I am thinking 150-5% = 142.50 + 5.50 = $148

3 0

4 years ago

What is the point-slope equation of the line with slope

Rainbow [258]

Answer:

y-5 = 3/2(x+3)

Step-by-step explanation:

The point slope form of the equation for a line is

y-y1 = m(x-x1) where m is the slope and (x1,y1) is a point on the line

y-5 = 3/2(x--3)

y-5 = 3/2(x+3)

3 0

3 years ago

Read 2 more answers

) 38 is what percent of 56?

goldfiish [28.3K]

Answer:

(38 ÷ 56) × 100% = 67.8%

hope this helps

6 0

3 years ago