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3 years ago

rewrite the standard form equation below into slope-intercept form. x-2y=3 A. y=x+3 By=-3/2x-1/2 C.y=3/2x-1/2 D.y=1/2x-3/2

1 answer:

8 0

Let us start converting!

$x-2y=3 \ \textgreater \ \mathrm{Subtract\:}x\mathrm{\:from\:both\:sides} \ \textgreater \ x-2y-x=3-x$

$-2y=3-x \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-2 \ \textgreater \ \frac{-2y}{-2}=\frac{3}{-2}-\frac{x}{-2}$

$\frac{-2y}{-2} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}:\ \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{2y}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ y$

$\frac{3}{-2}-\frac{x}{-2} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{3-x}{-2}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}\ \textgreater \ -\frac{-x+3}{2} \ \textgreater \ y = -\frac{3-x}{2}$

$Therefore... y=\frac{1}{2}x-\frac{3}{2} (Note: m=\frac{1}{2},\:b=-\frac{3}{2})$

M is the slope and b is the y intercept.

Hope this helps!

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LiRa [457]

We know: The total number of tickets is 35. Another way of saying this is the number of adults plus the number of children equals 35. i.e.:

$Adults+Children=35$

Using the symbols provided we can write this is as:

$a+c=35$

We also know: The total money collected is $75. We know that $6 was collected for every adult or $6a$ and $1 for every child or $1c$ . i.e.:

$(SixDollars \times Adults)+(OneDollar \ times Children)= $70$

Using the symbols provided and adding it all up we can write this is as:

$6a+1c=70$

Equation 1: $a+c=35$
Equation 2: $6a+c=70$

5 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 $\frac{dP}{dt}$

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

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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DaniilM [7]

Answer: if your question was C-A then it would be 2.42

Step-by-step explanation:

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3 years ago

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RSB [31]

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Whitepunk [10]

Answer:

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