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3 years ago

What is the perimeter?

Mathematics

1 answer:

tamaranim1 [39]3 years ago

8 0

Answer:

21.3 unit

Step-by-step explanation:

$side CB=\sqrt{(7^{2})+(2^{2})} =\sqrt{53}\\side CA=\sqrt{(5)^{2}+(2^{2})} =\sqrt{29} \\side AB=\sqrt{(5^{2} )+(7^{2} )} =\sqrt{74}$

$Perimeter=\sqrt{53} +\sqrt{29} + \sqrt{74} =21.2676$ unit

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Boating: A paddleboat moves at a rate of 12 mph in still water. If the rover’s current moves at a rate of 3 mph, how long will i

Lostsunrise [7]

The time taken by the boat to travel 32 miles downstream is 2.1 hours and for upstream 3.56 hours.

<h3>What is speed?</h3>

Speed is defined as the ratio of the time distance traveled by the body to the time taken by the body to cover the distance.

Given that:-

A paddleboat moves at a rate of 12 mph in still water. If the rover’s current moves at a rate of 3 mph.
how long will it take the boat to travel 36 miles downstream and 36 miles upstream?

The time taken by the boat upstream will be calculated as:-

( B + W ) Td = D

( 12 + 3 ) Td = 32

( 15 ) Td = 32

Td = 32 / 15 = 2.1 hours

The time taken by the boat downstream will be calculated as:-

( B - W ) Tu = D

( 12 - 3 ) Tu = 32

9 Tu = 32

Tu = 32 / 9 = 3.56 hours.

Therefore the time taken by the boat to travel 32 miles downstream is 2.1 hours and for upstream 3.56 hours.

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4 0

2 years ago

A closing is March 5 for a rental property with an annual tax bill of $2800. Calculate and show the debit and credits for the ta

kicyunya [14]

Answer:

The debit and credits for the tax proration will be as follows:

Debit seller for $483.29; and Credit buyer for $483.29.

Step-by-step explanation:

The assignment of how much is owed to the responsible party is the major reason of a proration.

For the days owned by the seller, the buyer needs money from the seller since the buyer will pay the taxes at end of the year.

Amount per day = Annual tax bill / 365 = $2800 / 365 = $7.67

Total number of days from January 1 to a day before March 5 = Number of days in January + Number of days in February + Number of days from March 1 to March 4 = 31 + 28 + 4 = 63

Amount the seller owes for the time he owned = Amount per day * Total number of days from January 1 to a day before March 5 = $7.67 * 63 = $483.29

Therefore, the debit and credits for the tax proration will be as follows:

Debit seller for $483.29; and Credit buyer for $483.29.

8 0

3 years ago

Solve the inequality: n - 7 > -4

ollegr [7]

Answer:

n-7>-4

n-7+7>-4+7

n>-4+7

n>3

3 0

3 years ago

Write an equation of the line that passes through the given points.<br> (-1,7) and (4, -8)

djverab [1.8K]

Answer:

y=-3x+4

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-8-7)/(4-(-1))

m=-15/(4+1)

m=-15/5

m=-3

y-y1=m(x-x1)

y-7=-3(x-(-1))

y-7=-3(x+1)

y=-3x-3+7

y=-3x+4

3 0

3 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago