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3 years ago

What is the answer 1 - 0.5137

Mathematics

2 answers:

4vir4ik [10]3 years ago

6 0

Answer:

0.4863

Step-by-step explanation:

Vladimir79 [104]3 years ago

4 0

0.4863 is the answer

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I need help with the question

Katena32 [7]

Answer:

6, 8, 4

Step-by-step explanation:

the answers are as follows

a. 6

b. 8

c. 4

3 0

2 years ago

Solve the equation of exponential decay.

BabaBlast [244]

Answer:

$9,220,000(0.888)^t

Step-by-step explanation:

Model this using the following formula:

Value = (Present Value)*(1 - rate of decay)^(number of years)

Here, Value after t years = $9,220,000(1 -0.112)^t

Value after t years = $9,220,000(0.888)^t

3 0

3 years ago

Complete the square and put this function in vertex form: f(x)=x^2+20x+97

Fofino [41]

Answer:

Step-by-step explanation:

f(x)=(x²+20x+100) -3 because : 97 = 100 - 3

f(x) = (x+10)² -3 ....vertex form

3 0

3 years ago

after allowing 15% discount on the price of an article and 13% bat was live and the remaining amount in the price of article bec

Eduardwww [97]

Answer:

14000

Explaination:

spwithvat = sp without vat + 13% of sp without vat

let, sp without vat be x.

sp with vat= x + 13% of x.

or, 13447 = x + 13/100 × x

or, 1344700 = 113x

or, x= 1344700/113

x = 11900

Let, mp be y.

sp with vat = mp - 15% of mp

or, 11900 = y - 15/100 × y

or, 1190000 = 85y

or, y= 1190000/85

y= 14000

5 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago