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3 years ago

Write y = 1/6x + 4 in standard form using integers.

2 answers:

8 0

The standard form of an equation of a line would be written as Ax + By = C where A, B and C are numbers. To write the given equation to this form, we do as follows:

<span>(y = x/6 + 4) 6
6y = x + 24
-x + 6y = 24

Therefore, the correct answer from the choices listed above is option B.

Hope this helps. Have a nice day.</span>

Nastasia [14]3 years ago

4 0

Answer:

-x + 6y = 24

Step-by-step explanation:

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I need some help solving this.

Ne4ueva [31]

Answer:

3+4+3+4+3+4+3+4=28

Step-by-step explanation:

the permiter is 28in

3 0

3 years ago

Read 2 more answers

Find the values of the variables

chubhunter [2.5K]

Y = 168° [ Because opposite angles are equal ]

x = 5x - 48 [ Because opposite angles are equal ]

4x = 48

x = 12°

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3 years ago

The surface area of a cube is 6 square centimeters. What is its volume, in cubic<br> centimeters?

Elenna [48]

Answer:

1 cubic centimeter is the volume

Step-by-step explanation:

the surface area is 6

6 = 6a^2

1 = a^2

1 = a

L * W * H = ?

1 x 1 x 1 = 1

1 is the volume

have a good day!

3 0

3 years ago

Iris is making hats for the members of the school marching band. She can make 3 hats in 1 1/2 hours. She wants to know how many

likoan [24]

She makes 3 hats per 1 and 1/2 hours. Therefore she makes $\frac{3}{\frac{3}{2}}$ hats per hour. This fraction simplified is 2 hats per hour. This is the solution.

2 hats per hour.

7 0

3 years ago

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2

Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

$y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\$

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t = e^−2(0) = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

$\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt} = -2te^{-2t} = -2(0)e^{-2(0)} = 0$

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

6 0

3 years ago