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alexandr1967 [171]
4 years ago
7

Line with y-intercept -4 through (6,0) y=.... PLEASE HELP

Mathematics
2 answers:
-Dominant- [34]4 years ago
8 0
Answer:  " y = \frac{2}{3} x − 4 " .

____________________________________
Note:  y = mx + b ;

 b = -4  ;  

So, we have:

y = mx - 4 ;

   Given the point on the line:  " (6, 0) " ;

So, when x = 6, y = 0 .
_____________________
y = mx - 4 ;  Solve for "m" ; So we can write the equation of the line, in slope-intercept format; that is:  "y = mx + b" ;

0  = m(6) - 4 ;

Add "4" to EACH SIDE of the equation ;

0 + 4 = m(6) - 4 + 4 ;

to get:

4 = 6m ;

Divide each side of the equation by "6" ;

4/6 = 6m / 6 ; 

to get:  2/3 = m ;

Now, we can rewrite the equation:

y = mx - 4 ;

y = \frac{2}{3} x − 4 .
________________________________________________________



SCORPION-xisa [38]4 years ago
5 0
2 points on the line are (0,-4) and (6,0)
slope Is thus 2,/3

y= 2/3 x - 4
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-2

Step-by-step explanation:

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Kitty [74]

A straight line is 180°. The angles of QRT (2x + 33) and TRS (5x - 21) make up the 180° angle. So you can do:

∠QRT + ∠TRS = 180°

(2x + 33) + (5x - 21) = 180°


2x + 33 + 5x - 21 = 180 Combine like terms

7x + 12 = 180   Subtract 12 on both sides

7x = 168   Divide 7 on both sides

x = 24


Now that you found x, you can find the angles of QRT and TRS.

m∠QRT = 2x + 33

           = 2(24) + 33

           = 48 + 33

∠QRT = 81°


m∠TRS = 5x - 21

          = 5(24) - 21

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g A cannonball is shot with an initial speed of 62 meters per second at a launch angle of 25 degrees toward a castle wall that i
IceJOKER [234]

Answer:

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Step-by-step explanation:

Given;

Initial speed v = 62m/s

Angle ∅ = 25°

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Height of wall y = 20

Resolving the initial speed to vertical and horizontal components;

Horizontal vx = vcos∅ = 62cos25°

Vertical vy = vsin∅ = 62cos25°

The time taken for the cannon ball to reach the wall is;

Time t = horizontal distance/horizontal speed

t = d/vx (since horizontal speed is constant)

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Applying the equation of motion;

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Substituting the given values;

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8 0
4 years ago
Help please hurry......
user100 [1]

All multiples of 15 between 0 and 300,inclusive

How?

\\ \sf\longmapsto f(x)=15x

\\ \sf\longmapsto f(0)=15(0)=0

\\ \sf\longmapsto f(20)=15(20)=300

Hence.

\\ \sf\longmapsto 0\leqslant R_f\leqslant 300

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