Please!!!! Order the following from least to greatest!!!

C 4. What is the simplified form of 345e 15c ?<br>b) 6c<br>9 se' 43<br>15c 5<br>4) 12a

levacccp [35]

I think it is b but no positive

8 0

3 years ago

Find x, y, and z. plsssssssssssssssssss

Alona [7]

Answer:

z= 104

y=76

x=76

Step-by-step explanation:

z+76= 180. solve for z

z=x

y=76

3 0

2 years ago

Carly purchased 9.5 pints of ice cream for a party. If each guest will be served exactly 3/5 pints of ice cream, what is the gre

aivan3 [116]

Answer:

15.8333333333

Step-by-step explanation:

9.5 divied by 3.5

3 0

3 years ago

Read 2 more answers

Mary loves eating fruits.Mary paid $7.98 for each of 2 berries and $ 5.99 for 3 peaches with 2 $20 dollars.How much change did M

kupik [55]

Answer:

$6.07$

Step-by-step explanation:

We know that Mary has two twenty dollar bills, equaling to forty dollars in total.

So lets start on the first few things she buys, the berries, each costing 7.98 dollars and cents, since she only bought two, that means we need to multiply 7.98 by 2

==> 15.96 dollars and cents

Now we can take it out of forty.

$ 40.00

- $ 15.96

-------------------

$24.04

Now we only have 24.04 dollars and cents.

Now for the next few items, the peaches which costs 5.99 each, then since he only bought 3, we multiply 5.99 three times.

==> 17.97 dollars and cents

Now we can take it out of 24.04.

$24.04

- $ 17.97

--------------------

$6.07

Mary has 6 dollars and 7 cents in change.

5 0

4 years ago

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

4 years ago