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2 years ago

Solve P = Check P-2w 2 = 1 for P.

Mathematics

1 answer:

GaryK [48]2 years ago

4 0

Answer:

P= 2l + 2W

Step-by-step explanation:

(P-2W)/2= l

(P-2W)= 2l

P-2W= 2l

P= 2l + 2W

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Which point gives the vertex of ƒ(x) = –x2 + 4x – 3?

Dmitriy789 [7]

Answer:

The vertex is (2,1)

Step-by-step explanation:

ƒ(x) = –x^2 + 4x – 3

Factor out the negative

= -(x^2 -4x+3)

Factor

What 2 numbers multiply to +3 and add to -4

-3*-1 = 3

-3+-1 = -4

f(x) = -( x-3)(x-1)

Find the zeros

0 = -( x-3)(x-1)

0 = x-3 0 = x-1

x=3 x=1

The x value of the vertex is 1/2 way between the two zeros

(3+1)/2 = 4/2 =2

To find the y value, substitute x=2 in

f(2) = -( 2-3)(2-1)

=-(-1)(1) = 1

The vertex is (2,1)

4 0

3 years ago

Five point six times four point eight

LenaWriter [7]

Answer:

26.88

Step-by-step explanation:start eight times four is 32

and than eight times six is 48 than add a zero and do four times six is 24

and four times five is twenty add it up and line up the decimal points

hope i helped !!

3 0

3 years ago

5x+14<4 or -2(x-8)<-2

Kitty [74]

Answer:

Step-by-step explanation:

5x+14<4 or -2(x-8)<-2

-14 -14 -2x+16<-2

5x<-10 -16 -16

÷5 ÷5 -2x<-18

x<-2 ÷-2 ÷-2

x>9 (sign changes because you divide by a negative number)

x<-2 or x>9

7 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001