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3 years ago

Nancy withdrew $19 from her bank account to pay for markers write in integer to represent this situation

Mathematics

2 answers:

lions [1.4K]3 years ago

7 0

Multiply 29 and 0.85 to get $24.65

Then, subtract 29- 24.65

Final answer: $4.35

ziro4ka [17]3 years ago

5 0

Hello!

Withdrawing is like taking out money from her account.

So, the integer that represents this situation is -$19.

I hope that was helpful! c:

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Graph: y < 1/3x+1/2

soldier1979 [14.2K]

Answer:

(-1.5,0) (0,0.5)

Step-by-step explanation:

I graphed it

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3 years ago

En una asamblea del colegio acuden padres de los cuatro cursos de la ESO. La cuarta parte de los padres son de 1º ESO, la décima

Evgen [1.6K]

Answer:

Parents from the four ESO courses attend a school assembly. A quarter of the parents are from 1st ESO, a tenth are from 2nd ESO, a quarter are from 3rd ESO and 56 parents are from 4th ESO. a) How many parents attended the assembly? b) How many parents from 1ºESO attended the assembly? c) How many parents of 2nd ESO attended the assembly? d) How many 3rd grade parents attended the assembly?

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP IM SO STUCK 20 POINTS

prohojiy [21]

Answer: X = 2

Step-by-step explanation: The correct answer to this question is that x=2. Regardless of what point you go to on the line, it always has a value of 2 because it is going vertically (up and down) at 2.

5 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

3 years ago

PLS HELP, I'm not very good at math so I need the answer to this

statuscvo [17]

Answer:

A BT = CT

Step-by-step explanation:

BAT ≅ CAT

That means

The angles are the same and the sides are the same by CPCTC

AB = AC

CT = BT

AT=AT

and

< BAT = <CAT

< ATB = <ATC

< TBA = <TCA

Given the choices on the left

A BT = CT is one of them

4 0

3 years ago