Hello could anyone tell me what 12 plus 14 times 2000 divided by 13 is?

Mathematics

1 answer:

sveticcg [70]2 years ago

5 0

The exact answer is 2,165.846153846154
Hope this helpssss :)

You might be interested in

Louann and Carla received equal scores on a test made up of multiple choice questions and an essay. Louann got 14 multiple choic

Gnoma [55]

Louann= 14x+24
Carla= 16x+14
So basically this is just like solving equations. The way I did this was just put random numbers for x and then solve it.
First attempt:
14(7)+24=16(7)+14
98+24=112+14
122=126
Okay the first one didn't work so I use 6 for x.
Second attempt:
14(6)+24=16(6)+14
84+24=96+14
108=110

So 6 didn't work and you already should know that the answers should be between 1 and 5.

So now I going to do 5 for x
Third attempt:
14(5)+24=16(5)+14
70+24=80=14
94=94

And so the answer is 5! :D The multiply choice questions were worth 5 points and Louann and Carla had 94 for their scores! I hope this help you out and clear for you! :D

7 0

3 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 $\frac{dP}{dt}$

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