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2 years ago

What is the equation of the line with the same slope as the equation x-4y=3 and the y-intercept(0,-8)?

Mathematics

1 answer:

kogti [31]2 years ago

6 0

Answer:

Answer:y = 1/4x -8Step-by-step explanation:First find the slope of the linex-4y =3Subtract x from each sidex-4y -x = -x+3-4y = -x+3Divide each side by -4-4y/-4 = -x/-4 +3/-4y = 1/4 x -3/4This is in the form y= mx+b where the slope is m and the y intercept is bThe slope is 1/4We want the same slopey = 1/4x +bwe know the y intercept is -8y = 1/4x -8.

Step-by-step explanation:

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The time to fly between new york city and chicago is uniformly distributed with a minimum of 120 minutes and a maximum of 150 mi

Paul [167]

120+150=270
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2 years ago

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Answer:

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1 year ago

Factorise (2t-7)²-(5t-4)² with working pls

sertanlavr [38]

Answer:

33+12t−21t^2

Step-by-step explanation:

(2t-7)²-(5t-4)²

Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².

4t^2−28t+49−(5t-4)²

Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².

4t^2−28t+49−(25t^2−40t+16)

To find the opposite of 25t^2

−40t+16, find the opposite of each term.

4t^2−28t+49−25t^2−40t+16

Combine 4t^2 and −25t^2 to get −21t^2.

−21t^2−28t+49+40t−16

Combine −28t and 40t to get 12t.

−21t^2+12t+49−16

Subtract 16 from 49 to get 33.

−21t^2+12t+33

Swap terms to the left side.

33+12t−21t^2

I hope this helped!

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2 years ago

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2 years ago

At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

Nikitich [7]

If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

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2 years ago