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Orlov [11]
3 years ago
12

Select the equations of the lines that are parallel to the line whose equation is y = 3x + 5.

Mathematics
1 answer:
Orlov [11]3 years ago
3 0
The slope-intercept form of the equation of a straight line:

y=mx+b

m=slope
Therefore; in this equation (y=3x+5); m=3

a)
6x+2y=12
2y=-6x+12
y=-6/2 x + 12/2
y=-3x+6  ⇒            m=-3≠3  (this line is not parallel to "y=3x+5")

b)
-3x+y=8
y=3x+8      ⇒          m=3 (this line es parallel to "y=3x+5")    

c)
3y=9x
y=9/3 x
y=3x          ⇒            m=3 (this line is parallel to "y=3x+5")

d)
y=-x            ⇒           m=-1≠3 (this line is not parallet to "y=3x+5")

Therefore; "-3x+y=8" and "3y=9x" are parallel to "y=3x+5".
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The sum of 2/3 and twice a number is equal to 5/6 subtracted from three times a number​
strojnjashka [21]

Answer:

3÷2

Step-by-step explanation:

First let the required no. be 'x'.

By question,

(2÷3)+2x=3x-(5÷6)

or,(2+6x)÷3=(18x-5)÷6

or,2+6x=(18x-5)÷2

or,4+12x=18-5

or,x=9÷6

Therefore, x=3÷2

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ira [324]

The Answer Is g(x)=1/3x^2

Use (3,3) to find the equation.

3^2x1/3=3

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A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long
stepladder [879]

Answer:  1.4 seconds

<u>Step-by-step explanation:</u>

The equation is: h(t) = at² + v₀t + h₀          where

  • a is the acceleration (in this case it is gravity)
  • v₀ is the initial velocity
  • h₀ is the initial height

Given:  

  • a = -9.81 (if it wasn't given in your textbook, you can look it up)
  • v₀ = 12
  • h₀ = 3

Since we are trying to find out when it lands on the ground, h(t) = 0

EQUATION:   0 = 9.81t² + 12t + 3

Use the quadratic equation to find the x-intercepts

                        a=-9.81, b=12, c=3

x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-9.81)(3)}}{2(-9.81)}\\\\\\x=\dfrac{-12\pm 16.2}{-19.62}\\\\\\x=\dfrac{-12+ 16.2}{-19.62}=-0.2\qquad x=\dfrac{-12- 16.2}{-19.62}=\large\boxed{1.4}\\

Note: Negative time (-0.2) is not valid

4 0
3 years ago
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