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IRISSAK [1]
3 years ago
7

Use the net price rate to calculate the net price for 10 boxes of computer paper if the unit price is $13.59 and a single trade

discount rate of 20% is allowed
Mathematics
1 answer:
torisob [31]3 years ago
5 0
$14.06, we can find the total cost by multiplying that by 30, to giveus $421.80 if there were no discount. We can find the discount price in two equivalent ways, either by finding 30% of 421.80 and subtracting that,or by finding 100% - 30% = 70%. That is, the new price will be 70% of the old price. Since they wantthe new price, and not the discounted amount, it is simpler to use 70% 70% = 70/100 or 0.7. 0.7 x 421.80 = $295.26
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FrozenT [24]
That is way too much questions for one sitting. If you really struggle that much, download this app called photomath. You take a picture of one of the questions and then it instantly solves it and shows the work so you know how to do it.

For 16-25 however, you would need to rewrite your questions on a blank piece of paper with the variables substituted in. So let's say:

a + b² - (c-2).

You would need to rewrite it as the following in order for the app to solve it.

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7 0
3 years ago
Write the equation of the line that passes through the points (−7,0) and (6,−4).
Over [174]

Answer:

y=-4/13x-28/13

Step-by-step explanation:

m=y2-y1/x2-x1

m=-4-0/6-(-7)

m=-4-0/6+7

m=-4/13

y-y1=m(x-x1)

y-0=-4/13(x-(-7))

y-0=-4/13x-28/13

y=-4/13x-28/13

5 0
3 years ago
Solve the equation 2x + 4 - 9 Explain the steps<br> and properties you used.
Verizon [17]

Answer:

2x - 5

Step-by-step explanation:

2x + 4 - 9

the 2x gonna stay

so you take 4 - 9 = -5

so the answers gonna be 2x - 5

bc the 5 is negative so you gonna subtract

I hope you understand my english...

6 0
3 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
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