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3 years ago

8 divided by 6.403 round to the nearest tenths

Mathematics

2 answers:

lions [1.4K]3 years ago

6 0

I believe the answer is 1.2

Mashutka [201]3 years ago

3 0

1.2 is the rounded answer

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Divide.

STatiana [176]

Answer:

189

Step-by-step explanation:

Step 1: the output value is 105

x=1.8% 105=100 (1) x=1.8% (2)

4 0

2 years ago

Please answer the question in the picture below. Thanks!

Arada [10]

Answer:

Yes

Step-by-step explanation:

$\frac{DH}{HE}=\frac{18}{36}=\frac{1}{2}$

$\text{This means that } DH=\frac{1}{2}HE$

and thus the given segments are parallel by the converse of the triangle proportionality theorem.

4 0

2 years ago

PLS HELP WILL MAKE FIRST RIGHT ANSWER BRAINLIEST

Amanda [17]

Answer: -8 is the answer.

Step-by-step explanation:

according to the question:

{{(8+4)*5}/(-10)}-2

= {[12*5}/(-10)]-2

= [60/(-10)]-2

= -6-2 = -8

4 0

2 years ago

Read 2 more answers

Which numbers are the extremes of the proportion 3/4=6/8

ki77a [65]

Answer:it’s 3 and 8

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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