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3 years ago

What is 10 2/1 divided by 2 1/4

Mathematics

2 answers:

stepladder [879]3 years ago

6 0

Answer:

5.3 but the three is repeated

Step-by-step explanation:

stealth61 [152]3 years ago

6 0

Answer:

$10 \frac{2}{1} \div 2 \frac{1}{4 } = \frac{48}{9}$

Penjelasan

$10 \frac{2}{1} = \frac{12}{1}$

$2 \frac{1}{4} = \frac{9}{4}$

$\frac{12}{1} \div \frac{9}{4} = \frac{12}{1} \times \frac{4}{9} = \frac{48}{9}$

Then the answer is <u>48/9</u>

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Valentin [98]

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3 years ago

Question 1 (1 point)

Naily [24]

Answer:

P ( -1, -3)

Step-by-step explanation:

Given ratio is AP : PB = 3 : 2 = m : n and points A(5,6) B(-5,-9)

We will calculate coordinates of the point P which divides line segment AB in the following way:

xp = (n · xa + m · xb) / (m+n) = (2 · 5 + 3 · (-5)) / (3+2) = (10-15) / 5 = -5/5 = -1

xp = -1

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8 0

3 years ago

X²-12x+a=(x+b)² find a and b

Valentin [98]

Answer:

Step-by-step explanation:

x² - 12x + a = (x + b)²

x² - 12x + a = x² + 2xb + b²

-12 = 2b and a = b²

b = -6

a = b² = 36

8 0

2 years ago

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

URGENT!!!!!

hichkok12 [17]

Answer:

Step-by-step explanation:

1. Cross Multiply. $\frac{5}{6} =\frac{30}{x}$

2. Solve. $\frac{5}{6} =\frac{30}{x} =36$

3. Your answer is 36

3 0

3 years ago

Read 2 more answers