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3 years ago

Compute the most rational way.

Mathematics

1 answer:

liraira [26]3 years ago

3 0

1.=9900
2.=9.3
3.= x= 12.2155
Hope this helps

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Fully simplify using only positive exponents.

Assoli18 [71]

Answer:

$\frac{xy^{3}}{25}$

Step-by-step explanation:

Given

$\frac{5x^6y^6}{125x^5y^3}$

Required

Simplify the given expression

$\frac{5x^6y^6}{125x^5y^3}$

Start by applying law of indices

$\frac{5x^{6-5}y^{6-3}}{125}$

$\frac{5xy^{3}}{125}$

Divide numerator and denominator by 5

$\frac{xy^{3}}{25}$

Hence; the expression is equivalent to $\frac{xy^{3}}{25}$

6 0

3 years ago

Square root of 6 - square root of 6 + square root of 6

Arte-miy333 [17]

Answer:

The answer is 2.44948974278 or 2.4.

Step-by-step explanation:

Hope this helps. Good luck!

4 0

3 years ago

10 Points!

Alex_Xolod [135]

Answer: 15

2 correct= 10+10=20
1 wrong= 20-5= 15

6 0

3 years ago

Read 2 more answers

This proportion can be solved to find the unknown length, x, in the smaller triangle. 124=6x use the drop-down menus to complete

Elodia [21]

The unknown X equals 20.66.

4 0

3 years ago

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A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago