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3 years ago

14

Find the circumference of the circle. Use 3.14 for pi

2 answers:

IgorLugansk [536]3 years ago

6 0

Answer:

The circumfrence is 14

Step-by-step explanation:

blondinia [14]3 years ago

5 0

The answer is 21.98

Explanation

3.14 x 7

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Nate made 8 loaves of bread on Monday and 10 more on Tuesday. He gave an equal number to each of his 9 neighbors. How many loave

mars1129 [50]

Answer:

2

Step-by-step explanation:

8 + 10 = 18

18/9 = 2

8 0

3 years ago

kari74 [83]

This expression is a perfect square trinomial. It has two answers but they don't have to be different. In this case it has to be the same - because it is a perfect square trinomial.

6 0

2 years ago

Read 2 more answers

a water wheel has a radius of 18 feet. The wheel is rotating at 10 revolutions per minute. Find the linear speed, in feet per

Yuri [45]

Answer:

The linear speed is $1,130.4\frac{ft}{min}$

Step-by-step explanation:

step 1

Find the circumference of the circle

The circumference is equal to

$C=2\pi r$

we have

$r=18\ ft$

$\pi =3.14$

substitute

$C=2(3.14)(18)$

$C=113.04\ ft$

step 2

Find the linear speed in feet per minute

we know that

The wheel is rotating at 10 revolutions per minute

Convert revolutions to feet

Remember that

One revolution is equal to the circumference of the circle

so

10 revolutions is equal to

$(10)113.04=1,130.4\ ft$

therefore

The linear speed is $1,130.4\frac{ft}{min}$

3 0

3 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

2 years ago

WARNING : NON SENSE ANSWER REPORT TO MODERATOR

Rzqust [24]

Answer:

$\frac{180}{147}$

Step-by-step explanation:

Simplify $(\frac{3}{-7} - \frac{11}{21} )$

$=> \frac{(-3 \times 3) - 11}{21}$

$=> \frac{-9 - 11}{21}$

$=> \frac{-20}{21}$

Find the additive inverse of $\frac{-20}{21}$ by using its property - <em>"Sum of a number & its additive inverse is always zero". </em>Assume that 'x' is an additive inverse of $\frac{-20}{21}$ .

$=> x + \frac{-20}{21} = 0$

$=> x = 0 - (-\frac{20}{21}) = \frac{20}{21}$

Simplify $(\frac{9}{5} \div \frac{7}{5} )$

$=> \frac{9}{5} \times \frac{1}{\frac{7}{5} }$

$=> \frac{9}{5} \times \frac{5}{7}$

$=> \frac{9}{7}$

Now, find the product of $\frac{9}{7}$ & $\frac{20}{21}$

$=> \frac{9}{7} \times \frac{20}{21}$

$=> \frac{180}{147}$

3 0

3 years ago