All categories

2 years ago

10 0 POINTS !! PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

Mathematics

1 answer:

Vitek1552 [10]2 years ago

8 0

Pythagorean Formula: c^2 = b^2 + a^2

$c^{2}$ = $3^{2}$ + $3^{2}$

$c^{2}$ = 9 + 9

$\sqrt{c^2$ = $\sqrt{18}$

c = 4.242640687 ft - 4.2 ft (Rounded to Nearest Tenth).

You might be interested in

A rectangular swimming pool measures 50 meters in length and 25 meters in width. A scale drawing of the swimming pool will be ma

igomit [66]

Answer:

the pool would be 10 cm by 5 cm

7 0

2 years ago

Read 2 more answers

PLEASE answer! Will give thanks (profile too) and 5-star rating (for sure) BRAINLIEST TOO if I have the option. Please answer. I

Otrada [13]

Answer:

148

Step-by-step explanation:

4 0

1 year 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

2 years ago

Write 24.652 as a mixed number.

Vladimir79 [104]

24.652 = 24 65/100

24 65/100 is your answer

hope this helps

4 0

2 years ago

Please help due tomorrow

mel-nik [20]

Answer: x= 2.5, y = 10

Step-by-step explanation:

<u><em>I'm going to assume that these photocopies are proportional in relations to each other.</em></u>

If they're proportional, you can set up two proportions:

$1) \frac{x}{5} =\frac{3}{6} \\\\2) \frac{5}{y} =\frac{3}{6}$

And cross-multiply:

$1) 6x = 5*3 \\\\2) 3y = 5*6$

Then solved for x and y:

$1) 6x = 15\\x=\frac{15}{6} =\frac{5}{2} =2.5 \\\\2) 3y = 30\\y=\frac{30}{3} =10$

5 0

2 years ago