Jamal runs the bouncy house at a festival. The bouncy house can hold a maximum of 1200 pounds at one time. He

A basketball team wants to paint half of a free-throw circle grey. If the circumference of the free-throw circle is 30.77 feet,

Anton [14]

Answer:

37.69 (≈ 38 ft²)

Step-by-step explanation:

2πr = 30.77

r = 30.77 / (2 * 3.14) = 4.9

half of free-throw circle area: πr² / 2 = (3.14 * 4.9²) / 2 = 37.69 (≈ 38 ft²)

3 0

3 years ago

3 solutions to y=-5x+1

Luda [366]

Y = -5x + 1

x = 1 , y = -4
x = 2 , y = -9
x = 3 , y = -14

so the answers are
(1,-4)
(2,-9)
(3,-14)

7 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

2 years ago

an airplane is at an altitude of 1,200 feet. It begins to ascend at a rate of 925 feet per minute. How many minutes will it take

Oksanka [162]

You start by subtracting: 16000-1200= 14800
Then you divide 14800 by how many feet the plane rises per minute
14800/945=16 minutes
The answer is 16 minutes

3 0

3 years ago

Read 2 more answers

A small hard drive measured 0.7553 centimeters across round to the nearest tenth

Alborosie

Answer:

0.8

Step-by-step explanation:

7 is in the tenth place since there is a 5 in the hundreds we round up to 8.

hope this helped =3

4 0

3 years ago