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3 years ago

What is 6 1/6 as a Decimal Please!

Mathematics

2 answers:

grigory [225]3 years ago

7 0

6 1/6 = 6.166666666...well basically six goes on forever.

Hope this helps:)

sdas [7]3 years ago

3 0

6 1/6 as a decimal is 6.16666667

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the height of a tree was 8 feet. after a year, tree's height increased by 25%. use a bar diagram to find the new height of the t

marysya [2.9K]

8×.25=2
8-2=6
so what you would do is make a bar graph where the y axis would be years and the x axis being height

5 0

3 years ago

Teri incorrectly solved the system of equations below. Correctly identify the error Teri made. 2x - 10 y = -30 x - 5y = 20 Step

umka21 [38]

Answer:

See explanations below

Step-by-step explanation:

Given the system of equation

2x - 10 y = -30 ... 1

x - 5y = 20 ... 2

Step: From 2;

x = 20 + 5y ...3

Step 2: Substitute equation 3 into 1

2x - 10y = -30

2(5y+20) - 10y = -30

Open the bracket

10y + 40 - 10y = -30

10y - 10y + 40 = -30

0+40 = -30

40 = -30

The steps taken by Teri are correct all through. He didn't make any error whatsoever

3 0

2 years ago

A can do a piece of work in 3 days and B in 6 days. In how many days will they do it together?

Sladkaya [172]

Answer:

What is a can???????plsssssss

3 0

2 years ago

Read 2 more answers

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

Ill give you 28 points if you answer this!!

elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

because youe last number is 90

6 0

2 years ago