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4 years ago

A jogger ran at a rate of 5.4 miles per hour. find the joggers exact rate ,in feet per minute

Mathematics

1 answer:

Lemur [1.5K]4 years ago

4 0

The jogger moves 5.4 mph / 60 = 0.09 miles per minute.

The 0.09 miles have to be divided by the length of one step.

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The area of a rectangular park is 3/5 square mile. The length of the park is 7/8 mile. What is width of the park

Ira Lisetskai [31]

3/5 is 0.6
7/8 is 0.875
0.6/0.875 is 0.685714286
0.685714286 is 685714286/1000000000
(not so sure about it but that was what i came up with hope i helped)

6 0

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

Please Help. Find the area. Thank you

Andru [333]

Answer:

432 yd squared

Step-by-step explanation:

This is a parallelogram. The area of a parallelogram is denoted by: A = bh, where b is the base and h is the height. Here, the base is b = 19 1/5 and the height is h = 22 1/2. Plug these in:

A = bh = (19 1/5) * (22 1/2)

To make this simpler, let's convert the numbers into decimals. 1/5 is just 0.2 so 19 1/5 is 19.2. 1/2 is just 0.5, so 22 1/2 is 22.5. Now we have:

A = 19.2 * 22.5 = 432

Thus the area is 432 yd squared.

Hope this helps!

7 0

3 years ago

Read 2 more answers

The measure of the exterior angle of the triangle is what degree?

RSB [31]

Answer:

92°

Step-by-step explanation:

By exterior angle theorem:

$(2x - 2) \degree = x \degree + 45 \degree \\ \\ (2x - 2 - x) \degree = 45 \degree \\ \\ (x - 2) \degree = 45 \degree \\ \\ (x - 2) = 45 \\ \\ x = 45 + 2 \\ \\ x = 47 \\ \\ (2x - 2) \degree = (2 \times 47 - 2)\degree \\ \\ (2x - 2) \degree = (94 - 2)\degree \\ \\ (2x - 2) \degree = 92\degree \\ \\ measure \: of \: the \: exterior \: angle \\ = 92 \degree$

4 0

3 years ago

Anyone know the answers for #8 and #9?

babymother [125]

Check the picture below, notice those three quadrilaterals.

the square, is really a rectangle and is also a rhombus, but with right-angles and equal sides.

8 0

3 years ago