All categories

2 years ago

$123.74856 rounded to the nearest cent would be?

2 answers:

8 0

$123.74856 rounded to the nearest cent is $123.75. when your rounded you take the 3 number or the number in the thousadths place and if its over 5 you round up if its under five you leave it alone

Airida [17]2 years ago

3 0

$123.75 you would round all of them to the nearest hundreth

You might be interested in

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago

What are the solutions of the quadratic equation 4x2 - 8x – 12 = 0?

Juliette [100K]

Answer:

see bottom

Step-by-step explanation:

divide through by 4

x2 - 2x - 3 = 0

x2 - 3x + x - 3 = 0

(x2 - 3x) + (x - 3) = 0

x(x - 3) + 1(x - 3) = 0

(x + 1) (x - 3) = 0

x + 1 = 0 and x - 3 = 0

x = -1 and x = 3

5 0

2 years ago

If AC=3x, AB=15, and BC=2x-1, how could we solve for x? A. Set 3x=15 B. Set 3x=2x-1 C. Set 3x+15=2x-1 D. Set 3x=15+ (2x-1) E. No

k0ka [10]

Hmm this is a very challenging question good luck finding someone who can answer it lol

5 0

3 years ago

14 POINTS 7x - 3 = 18

Svetlanka [38]

Subtract 3 first so it would be 7x=15
then divide by 7 so x would be approximately 2.14

5 0

2 years ago

Read 2 more answers

7.<br>(-3x³)(-2x³y⁴z)(-3z²)÷(4x³z)(-3yz) (-3xyz)

serg [7]

Answer:

$-\frac{x^2y^2}{2}$

Step-by-step explanation:

We have an extremely large equation and are asked to divide it, so let's solve it step-by-step :

Remove the parenthesis to make it easier to read :

$\frac{-3x^3 *2x^3y^4z *3z^2}{4x^3z*3yz*3xyz}$

Multiply the numerators :

$\frac{-18x^6y^4z^3}{4*3*3x^3yyzzz}$

Multiply the denominators :

$\frac{-18x^6y^4z^3}{36x^4y^2z^3}$

Apply the negative rule :

$-\frac{18x^6y^4z^3}{36x^4y^2z^3}$

Cancel the common factor which is 18 :

$-\frac{x^6y^4z^3}{2x^4y^2z^3}$

Apply the addition exponent rule :

$\frac{y^4z^3x^{6-4} }{2y^2z^3}$

Subtract :

$\frac{x^2y^4z^3 }{2y^2z^3}$

Apply the rule for y :

$\frac{x^2y^{4-2} z^3 }{2z^3}$

Subtract :

$\frac{x^2y^2z^3 }{2z^3}$

Cancel the common factor of z^3 :

$-\frac{x^2y^2}{2}$

5 0

2 years ago