If a circle has a diameter of 7 feet what is it’s circumference

Answer ASAP I needs this like asap

jonny [76]

Answer:

1. 84 in

2. 300 ft

3. 2500 cm

4. 5000 m

5. 6 lbs

6. 10560.03937 ft

7. 3520.0131234 yd

8. 4.375 c

9. 1.5 L

10. 6000 mg

11. 48

12. 80000 lbs

13. 1.5 m

14. 1500 mm

15. Dan's phone is 157 millimeters long. How many inches is this? 5

Step-by-step explanation:

3 0

2 years ago

How many faces does the following shape have?

Allisa [31]

The shape has 4 faces . i hope this helps

7 0

2 years ago

Read 2 more answers

Erik's disabled sailboat is floating stationary 3 miles east and 2 miles north of Kingston. A ferry leaves Kingston heading towa

sergij07 [2.7K]

Answer:

Step-by-step explanation:

Let's suppose that Ballard is an origin with coordinates (0;0)

Ballard is 8 miles south and 1 mile west of Edmonds ⇒ Edmonds is 1 mile east and 8 miles north of Ballard.

Thus, coordinates of Edmonds (0+1; 0+8) (1;8)

Edmonds is 6 miles due east of Kingston. So, is 6 miles due of Edmonds

Thus, Kingston coordinates (1-6;8) (-5:8)

Sailboat is 3 miles east and 2 miles north of kingston

So, coordinates of sailboat are (-5+3; 8+2) (-2;10)

a)

Ferry leaves Kingston towards Edmonds at mph and Edmonds is 6 miles due east of Kingston.

Initial ferry coordinates as calculated above are equal to Kingston coordinates (-5;8)

After 20 minutes the ferry turns to south.

This distance travelled (20min) is d=9mph*20/60h=3 miles

So, ferry travelled 3 miles toward east in 20 min

Then, coordinates became (-5+3;8+0) (-2;8)

Thus, we have a line connecting two points (-5;8) and (-2;8)

Line connecting them has an equation:

y-8 = (8-8)/(-2+5) * (x+5)

y-8=0

y=8 - This is the equation for the first 20 minutes of travel

Then, ferry turns due south (-2;8) and has a vertiacl line

The equation of verical line is x=a, so the equation will be x=2

b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat.

region looks like a circular disc with a center (-2;10)

(x+2)^2+(y-10)^2=3^2

(x+2)^2+(y-10)^2 <9 (interior of the circular disc)

(x+2)^2+(y-10)^2 >9 (exterior)

The equation of the line joining the Kingston and Edmonds is y=8

the point of intersection:

(x+2)^2+(8-10)^2=9

(x+2)^2=5

x is approximately 0.24; -4.24

(0.24;8) (-4.24;8) - intersection points

c) The ferry exits the radar during the trid due south long x=-2

The points of intersection of the circle and the line x=-2:

(-2+2)^2+(y-10)^2=9

0+(y-10)^2=0

y is 7.13

(-2;7)

d) Takes south turn at (-2;8), then ferry travels 0.8 miles up to point (-2;7) where it exists the radar zone

Speed is 9mph

The time taken to cover 1 mile = t=1/9hr=1/9 *60=6.667min

So, it exists after 6.67 minutes

e) Ferry enters at (-4.24;8) and takes turn due to south at (-2;8)

The distance travelled is = [-4.24+2]=2.24 miles

The time taken to cover these miles is = t=2.24/9 *60=14.93 min

After turning to south ferry remains in radar for 6.7 min

So, it remains in radar zone for 14.93+6.7=21.63min

8 0

2 years ago

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability

$\dfrac{96}{\binom{52}5}\approx0.0000369$

and so the events A and B are NOT independent.

4 0

3 years ago

Find a quadratic equation with roots -1, 4i and -1 - 4i

vazorg [7]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }i^2=-1\\\\
-------------------------------\\\\
\begin{cases}
x=-1+4i\implies &x+1-4i=0\\
x=-1-4i\implies &x+1+4i=0
\end{cases}
\\\\\\
(x+1-4i)(x+1+4i)=\stackrel{y}{0}
\\\\\\\
[(x+1)~~-~~(4i)][(x+1)~~+~~(4i)]=y
\\\\\\\
[(x+1)^2~~-~~(4i)^2]=0\implies (x^2+2x+1)~-~(4^2i^2)=y
\\\\\\
(x^2+2x+1)~-~(16\cdot -1)=y\implies (x^2+2x+1)~-~(-16)=y
\\\\\\
x^2+2x+1~~+16=y\implies x^2+2x+17=y$

3 0

3 years ago