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3 years ago

Helpppppppppppppp!!!!!!!!!!!!!!!

Mathematics

2 answers:

Blababa [14]3 years ago

7 0

7/18 because 7 times 1 is 7 and 9 times 2 is 18 :)

amid [387]3 years ago

6 0

$\frac{7}{18}$

Good luck! (:

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Keenan will run 2.5 miles from his house to Jared’s house. He plans to hang out for 45 minutes before walking home. If he can ru

jenyasd209 [6]

Answer:

Kennan will be from home approximately an hour and 48 minutes.

Step-by-step explanation:

We must know that total time ( $t_{T}$ ) that Keenan will be from home is the sum of run ( $t_{R}$ ), hang out ( $t_{H}$ ) and walk times ( $t_{W}$ ), measured in hours:

$t_{T} = t_{R}+t_{H}+t_{W}$

If Keenan runs and walks at constant speed, then equation above can be expanded:

$t_{T} = \frac{x_{R}}{v_{R}}+t_{H}+ \frac{x_{W}}{v_{W}}$

Where:

$x_{R}$ , $x_{W}$ - Run and walk distances, measured in miles.

$v_{R}$ , $v_{W}$ - Run and walk speeds, measured in miles per hour.

Given that $x_{R}=x_{W} = 2.5\,mi$ , $v_{R} = 6\,\frac{mi}{h}$ , $v_{W} = 4\,\frac{mi}{h}$ and $t_{H} = 0.75\,h$ , the total time is:

$t_{T} = \frac{2.5\,mi}{6\,\frac{mi}{h} } + 0.75\,h+\frac{2.5\,mi}{4\,\frac{mi}{h} }$

$t_{T} = 1.792\,h$ ( $1\,h\,48\,m$ )

Kennan will be from home approximately an hour and 48 minutes.

7 0

3 years ago

HELPPPPPPPPPPPP

vlabodo [156]

Answer:

25cm

Step-by-step explanation:

one the possibility 36 - 11 =25 the answer is 25

8 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago

1. Which of the following is an example of a system of equations?

Anni [7]

Answer: (1)

Step-by-step explanation:

A system of equations must have two equations. The first option is the only one that satisfies this.

7 0

2 years ago

Sin2A + cos4A = cos2A + sin4A

xxMikexx [17]

Here is the graph i’ve created to answer your question!