I will mark brainlest and give 50 points.

atrell will rotate △abc 180° about one of the labeled points. latrell says, “the image will only touch the pre-image at ...

kompoz [17]

The labeled point in which it will be rotated around

8 0

2 years ago

What is the equation of the line that is parallel to the line 2x − y = 3 and passes through the point (2, −1)?

Ivanshal [37]

Answer:

C. y = 2x − 5

Step-by-step explanation:

Given equation of line

2x − y = 3

rewriting the equation in slope intercept form as answer are given slope intercept form

y = 2x - 3

slope intercept form of equation is

y = mx + c

where m is the slope

c is y intercept

thus

slope for y = 2x - 3 when comparing with y = mx + c is 2

now we know slope of two parallel line is same

thus,

the slope of line that is parallel to the line 2x − y = 3

is 2

Let the equation of required line be y = mx + c

where m = 2

thus

y = 2x + c is the new required equation

it passes through the point (2, −1)

using y = -1 and x = 2 in y = 2x + c

-1 = 2*2 + c

c = -1 - 4

c = -5

thus,

equation of required line is option c

y = 2x - 5 ----answer

7 0

2 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 $\frac{dP}{dt}$

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