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3 years ago

Two teachers purchase supplies for their classrooms.

Mathematics

1 answer:

inna [77]3 years ago

7 0

Answer:

Cost of One Journal = $3.99

Cost of One pen = $0.59

Step-by-step explanation:

Let

Cost of One Journal = x

Cost of One pen = y

We can make equation from given statements.

Mr. Bowden purchases 18 journals and 40 pencils for $95.42.

$18x+40y=95.42$

Ms. Jacinto purchases 11 journals and 16 pencils for $53.33.

$11x+16y=53.33$

Now solving these equations to find the value of x

$18x+40y=95.42--eq(1)\\11x+16y=53.33--eq(2)$

Multiply eq(1) with 2 and eq(2) with 5

$36x+80y=190.84\\55x+80y=266.65\\-\:\:\:\:-\:\:\:\;\:\:\:\:\:\:\:-\\---------\\-19x=-75.81\\x=\frac{-75.81}{-19}\\x=3.99\\$

We get the value of x is: x=3.99

Now, putting value of x in equation 1 to find value of y.

$18x+40y=95.42\\18(3.99)+40y=95.42\\71.82+40y=95.42\\40y=95.42-71.82\\40y=23.6\\y=\frac{23.6}{40}\\y= 0.59$

So, we get the value of y: y = 0.59

Now, finding the costs:

Cost of One Journal = x = $3.99

Cost of One pen = y = $0.59

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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 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

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$