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4 years ago

I REALY NEED HELP!

Mathematics

2 answers:

dalvyx [7]4 years ago

6 0

Since ACD is equal to 82, then DCE would equal 41. This is because it is half of 82.

podryga [215]4 years ago

6 0

Answer:

(A)41°.

Step-by-step explanation:

We know that in Kite, the diagonals are the angle bisectors, therefore

In Kite ABCD, m∠ACD=82°, thus

$m{\angle}DCE=\frac{1}{2}m{\angle}ACD$

Substituting the given value, we have

⇒ $m{\angle}DCE=\frac{1}{2}(82^{\circ})$

⇒ $m{\angle}DCE=41^{\circ}$

Therefore, the measure of m∠DCE is 41°.

Hence, option A is correct.

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Neveah has created the function of f(x) = 2x+4/3 to represent the growth of her hair, where x represents the number of months si

Andre45 [30]

F(x)=2x+4/3

Replace f(x) by y:
y=2x+4/3

Solving for x:
y-4/3=2x+4/3-4/3

Subtracting the two terms on the left side of the equation:
(3y-4)/3=2x

Dividing both sides of the equation by 2:
[(3y-4)/3]/2=2x/2
[(3y-4)/3]*(1/2)=x
[(3y-4)*1]/[(3)*(2)]=x
(3y-4)/6=x
x=(3y-4)/6

Replace "x" by "f^(-1) (x)" (inverse function) and "y" by "x":
f^(-1) (x) = (3x-4)/6 Inverse function

Then, for x=6
f^(-1) (6) = [ 3(6)-4]/6
f^(-1) (6) = (18-4)/6
f^(-1) (6)= 14/6
f^(-1) (6) = 7/3

Her hair will have grown 6 after 7/3 = 2.33 months

3 0

3 years ago

A tank with capactity 500 gal originally contains 200 gal water with 100 lbs of salt mixed into it. Water containing 1 lb of sal

prisoha [69]

Answer:

The amount of salt in the tank at any moment $t$ is

$f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

The concentration of salt in the tank when it is at the point of overflowing is $0.968$ .

The theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

Step-by-step explanation:

Let $f(t)$ be the amount of salt in the tank at any time $t.$

Then, its time rate of change, $f'(t)$ , by (balance law).

Since three gallons of salt water runs in the tank per minute, containing $1$ lb of salt, the salt rate is

$3.1=3$

The amount of water in the tank at any time $t$ is.

$200+(3-2)t=200+t,$

Now, the outflow is $2$ gal of the solution in a minute. That is $\frac 2{200+t}$ of the total solution content in the tank, hence $\frac 2{200+t}$ of the salt salt content $f(t)$ , that is $\frac{2f(t)}{200+t}$ .

Initially, the tank contains $100$ lb of salt,

Therefore we obtain the initial condition $f(0)=100$

Thus, the model is

$f'(t)=3-\frac{2f(t)}{200+t}, f(0)=100$

$\Rightarrow f'(t)+\frac{2}{200+t}f(t)=3, f(0)=100$

$p(t)=\frac{2}{200+t} \;\;\text{and} \;\;q(t)=3$ Linear ODE.

so, an integrating factor is

$e^{\int p dt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2$

and the general solution is

$f(t)(200+t)^2=\int q(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac 1{(200+t)^2}\int 3(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac c{(200+t)^2}+200+t$

Now using the initial condition and find the value of $c$ .

$100=f(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}$

$\Rightarrow c=-4000000=-4\times 10^6$

$\Rightarrow f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

is the amount of salt in the tank at any moment $t.$

Initially, the tank contains $200$ gal of water and the capacity of the tank is $500$ gal. This means that there is enough place for

$500-200=300$ gal

of water in the tank at the beginning. As concluded previously, we have one new gal in the tank at every minute. hence the tank will be full in $30$ min.