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3 years ago

M⁵ = what when m = ⅔

Mathematics

2 answers:

Greeley [361]3 years ago

5 0

Answer:

Step-by-step explanation:

M^5

(2/3)^5

2^5/3^5

32/243

Tanzania [10]3 years ago

3 0

Answer:

0.13

Step-by-step explanation:

M ^5

Divide 2/3

Equals 0.6667

0.6667 ^ 5

Equals 0.13 Rounded

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Convert this into standard form: y=(4/5)x-2

hjlf

Answer:

4x - 5y = -10

Step-by-step explanation:

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3 years ago

Can you help me Graph y= -x - 6?

Naya [18.7K]

Plot point(0,-6). Then go down 1 and 1 to the right locate another point(1,-7) connect (0,-6) and(1,-7) and extended from both sides

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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

Olive walks at a rate of 4 miles per hour. She lives 2 miles from the news station.How long would it take her to walk home from

Montano1993 [528]

4/1 = 2/x...4 miles to 1 hr = 2 miles to x hrs
cross multiply because this is a proportion
(4)(x) = (1)(2)
4x = 2
x = 2/4 reduces to 1/2...so she can walk 2 miles in 1/2 hr (or 30 minutes)

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Oduvanchick [21]

Answer:

Step-by-step explanation:

Because if your going to divide a whole the whole number needs to be 3/1 so it A the mistake

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3 years ago

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