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3 years ago

The vertices of a quadrilateral in the coordinate plane are known. How can the perimeter of the figure be found?

Mathematics

2 answers:

bonufazy [111]3 years ago

7 0

Use the distance formula to find the length of each side, and then add the lengths.

iren2701 [21]3 years ago

6 0

Answer:

Step-by-step explanation:

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Find the value of x.

jok3333 [9.3K]

Answer:

Complementary

Step-by-step explanation:

5 0

2 years ago

Item 16 An indoor water park has two giant buckets that slowly fill with 1000 gallons of water before dumping it on the people b

nignag [31]

Answer:

3:16 P.M.
5:22 P.M.

Step-by-step explanation:

The two dump intervals have a greatest common factor (GCF) of 3, so their least common multiple (LCM) is ...

(18)(21)/3 = 126 . . . . minutes

This period is 2 hours 6 minutes. The last time both dumped was 1:10, so the next time both will dump is ...

1:10 +2:06 = 3:16 . . . P.M.

and the next time after that is ...

3:16 +2:06 = 5:22 . . . P.M.

4 0

3 years ago

What are the endpoint coordinates for the midsegment of <br> △JKL that is parallel to line JL?

nirvana33 [79]

(2.5, 2.5) and (2, 0)
But you knew that already
sorry

3 0

3 years ago

Read 2 more answers

At what point on the graph of y=2e^X -1 is the tangent line parallel to the line y= 5x -1

IgorLugansk [536]

Paralell has same slope
y=mx+b
m=slope
y=5x-1
when is the slope 5?

take the derivitive of 2e^x
f'(2e^x)=2f'(e^x)=2e^x

that is the slope
5=2e^x
divide both sides by 2
2.5=e^x
take the ln of both sides
ln2.5=x

when x=ln2.5 they are paralell

5 0

3 years ago

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

3 years ago