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3 years ago

MORE EASY POINTS FOR Y'ALL

Mathematics

2 answers:

hodyreva [135]3 years ago

6 0

Answer:

False

Step-by-step explanation:

There are 2 y values for 6 and 12. A function only has one for each x value.

Hope I helped!!!!

sergejj [24]3 years ago

3 0

Answer: A

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Which ordered pair is a solution of the equation? -x-4y=-10

Makovka662 [10]

Answer:

(2, 2 )

Step-by-step explanation:

To find a solution, choose any value for x, substitute into the equation and solve for y.

Choose x = 2, then

- 2 - 4y = - 10 ( add 2 to both sides )

- 4y = - 8 ( divide both sides by - 4 )

y = 2

Thus (2, 2 ) is a solution to the equation

8 0

4 years ago

Read 2 more answers

(6z 2 - 4z + 1)(8 - 3z).

Yakvenalex [24]

Distributive property
a(b+c)+ab+ac
a(b-c)=ab-ac

(6z^2-4z+1)(8-3z)
move for nicety
(8-3z)(6z^2-4z+1)
distribute
8(6z^2-4z+1)-3z(6z^2-4z+1)=
48z^2-32z+8-18z^3+12z^2-3z=
-18z^3+48z^2+12z^2-32z-3z+8=
-18z^3+60z-35z+8

7 0

3 years ago

Is that the only ones I used

jek_recluse [69]

Answer:

yes

Step-by-step explanation:

7 0

3 years ago

The length.of a hypotenuse of a right triangle is 32feet and the length of one of the lrgs is 18 feet. what is the length, to th

gayaneshka [121]

Use Pythagorean's Theorem to solve for the missing leg length.

$32^2-18^2=a^2$

with a being the length of the missing leg. Simplifying we have

$1024-324=a^2$

and

$a^2=700$ . If we are told to express our answer to the nearest tenth of an inch, then we are simply plugging that into our calculator to get 26.45 or, rounded correctly, 26.5

6 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