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3 years ago

10

Please help me out with this!!!!!!!!!!!!

1 answer:

vesna_86 [32]3 years ago

6 0

Answer:

y = - $\frac{3}{4}$ x + 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 3) and (x₂, y₂ ) = (4, 0) ← 2points on the line

m = $\frac{0-3}{4-0}$ = - $\frac{3}{4}$

Note the line crosses the y- axis at (0, 3 ) ⇒ c = 3

y = - $\frac{3}{4}$ x + 3 ← equation of line

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2 years ago

HELP URGENT BRAINLIEST

nadya68 [22]

Given:

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To find:

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$\sin \theta=\dfrac{Opposite}{Hypotenuse}$

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2 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

7 0

3 years ago