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2 years ago

15

This negative and positive stuff is really confusing me can anyone please help?

1 answer:

RSB [31]2 years ago

7 0

Answer:

Positive integers have values greater than zero. Negative integers have values less than zero. Zero is neither positive nor negative. The rules of how to work with positive and negative numbers are important because you'll encounter them in daily life, such as in balancing a bank account, calculating weight, or preparing recipes. (THIS EXPLAINS WHT THEY R, HOPE THIS HELPS U)

Step-by-step explanation:

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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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