The area of circular garden is 530.66 feet2. What is the circumference of the garden? (Use 3.14 for .

1 answer:

3 0

$\bf \textit{area of a circle}\\\\
A=\pi r^2\quad 
\begin{cases}
r=radius\\
------\\
A=530.66
\end{cases}\implies 530.66=\pi r^2
\\\\\\
\cfrac{530.66}{\pi }=r^2\implies \sqrt{\cfrac{530.66}{\pi }}=r
\\\\\\
\textit{circumference of a circle}\\\\
C=2\pi r\qquad \implies C=2\pi \left( \sqrt{\cfrac{530.66}{\pi }} \right)$

You might be interested in

Round these decimals to 2 decimals places.Please I need some help...

Naddik [55]

A) 1.15
B) 2.87
C) 3.59
D) 4.22
E) 5.74
F) 6.95
G) 7.25
H) 8.64
I) 9.68
J) 28.33

8 0

3 years ago

Read 2 more answers

A plumber's daily earnings have a mean of $145 per day with a standard deviation of $16.50.

avanturin [10]

Step-by-step explanation:

The plumber's daily earnings have a mean of $145 per day with a standard deviation of

$16.50.

We want to find the probability that the plumber earns between $135 and

$175 on a given day, if the daily earnings follow a normal distribution.

That is we want to find P(135 <X<175).

Let us convert to z-scores using

$z = \frac{x - \mu}{ \sigma}$

This means that:

$P(135 \: < \: X \: < \: 175) = P( \frac{135 - 145}{16.5} \: < \: z \: < \frac{175 - 145}{ 16.5} )$

We simplify to get:

$P(135 \: < \: X \: < \: 175) = P( - 0.61\: < \: z \: < 1.82 )$

From the standard n normal distribution table,

P(z<1.82)=0.9656

P(z<-0.61)=0.2709

To find the area between the two z-scores, we subtract to obtain:

P(-0.61<z<1.82)=0.9656-0.2709=0.6947

This means that:

$P(135 \: < \: X \: < \: 175) =0.69$

The correct choice is C.

7 0

3 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$