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3 years ago

The y-intercept of the graph of a line is (-2). The line also passes through the point (5, 1).

Mathematics

1 answer:

MAVERICK [17]3 years ago

7 0

Answer:

y = 3/5 x - 2

Step-by-step explanation:

To find the equation of the line, calculate the slope using two points. With the slope and one point, substitute them into the point slope form and simplify as needed.

Use the points (5,1) and (0, -2) since -2 is the y-intercept.

Slope = $\frac{y_2-y_1}{x_2-x_1}=\frac{1--2}{5-0}=\frac{3}{5}$

y - -2 = 3/5 (x - 0)

y + 2 = 3/5(x)

y = 3/5 x - 2

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3 years ago

Read 2 more answers

Let θ

konstantin123 [22]

Answer:

θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

Step-by-step explanation:

Given :

Length of side opposite to angle θ is y

Length of side adjacent to angle θ is x

θ is part of a right angle triangle

At this instant,

x = 8 , $\frac{dx}{dt}$ = 7

( $\frac{dx}{dt}$ denotes the rate of change of x with respect to time)

y = 8 , $\frac{dy}{dt}$ = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ = $\frac{y}{x}$ -------------------------------------------------------------------1

At this instant,

tanθ = $\frac{8}{8}$ = 1

Therefore θ = π/4

We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or $\frac{d}{dt}$ (θ)

$\frac{d}{dt}$ (tanθ) = $\frac{d}{dt}$ (y/x)

( Applying chain rule of differentiation for R.H.S as y*1/x)

$sec^{2}$ θ $\frac{d}{dt}$ (θ) = $\frac{1}{x}$ $\frac{dy}{dt}$ - $\frac{y}{x*x}$ $\frac{dx}{dt}$ -----------------------2

Substituting the values of x , y , $\frac{dx}{dt}$ , $\frac{dy}{dt}$ , θ at that instant in equation (2)

2 $\frac{d}{dt}$ (θ) = $\frac{1}{8}$ *(-14)- $\frac{8}{8*8}$ *7

$\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

Therefore θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

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