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Lana71 [14]
3 years ago
12

Solve the system by graphing y = -3/4x + 6 y = 2x - 5

Mathematics
1 answer:
jonny [76]3 years ago
4 0

Answer:

X= 4. Y = 3.

Step-by-step explanation:

If you have a graphing calculator, put the first equation into Y1. And second equation into Y2. Then, find the intersection of the two equations. That's how you find your answer.

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Simplify: x(x-2)-3x(2x-1)
julsineya [31]

Answer:

-5x^2 +x

Step-by-step explanation:

x(x-2)-3x(2x-1)

Distribute

x^2 -2x -6x^2 +3x

Combine like terms

-5x^2 +x

4 0
3 years ago
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Given the table below find f(g(3))
12345 [234]

Answer:

Steps in provided image.

Step-by-step explanation:

5 0
3 years ago
Micheal has 3 quarters, 2 dimes, and 3 nickels
Anna35 [415]

Answer:

$1.10

explanation:

5 0
3 years ago
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The function f(x) = 200(0.5)x/50 models the amount in pounds of a particular
victus00 [196]

<u>Answer:</u>

29 years

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

  • Given the function f(x)= 200(0.5)^{\frac{x}{50} }, where x stands for the number of years since the material was put into the vault, we can plug 140 into x to find the amount leftover.
  • Another way of doing this is to change the function to f(x)=200(0.5^{\frac{1}{50} })^{x} which are equivalent functions, and then plug 140 into x. This function is helpful if you also needed to find the rate.
  • I'll use the 1st function because the question doesn't ask for the rate

f(x)= 200(0.5)^\frac{140}{50}}

From here I'll just use a calculator & get 28.7, or 29 years.

4 0
2 years ago
find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2
Alexxandr [17]

The surface is parameterized by

\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k

and the normal to the surface is given by the cross product of the partial derivatives of \vec s :

\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}

It looks like you're given

\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}

Then the normal vector is

\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k

Now, the point (2, 2, 0) corresponds to u and v such that

\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}

and solving gives u = v = 1, so the normal vector at the point we care about is

\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k

Then the equation of the tangent plane is

\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath +  (z-0)\,\vec k\right) = 0

-4(x-2) + 2(y-2) - 4z = 0

\boxed{2x - y + 2z = 2}

6 0
2 years ago
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