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Gnom [1K]
3 years ago
5

A line with a negative slope is perpendicular to one of the lines shown. What is its slope? Use the graph shown below. The Slope

is:

Mathematics
1 answer:
goblinko [34]3 years ago
7 0

Answer:

  • The slope for the red line is decreasing and hence the slope must be negative. -1x or simply -x
  • The slope of the blue line is increasing and hence positive: 1/2x
  • The solution to the equation is the intercept (2,3)
  • Rate positively and give brainlist
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180 - 64= 116

116 divided by 4 = 29

x=29

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5 0
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there are 5 green marbles, 4 red marbles, and 3 yellow marbles in a bag. John selects a marble at random, keeps it, and then sel
Daniel [21]

Answer:

1/22

Step-by-step explanation:

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3 0
2 years ago
Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).
Elan Coil [88]

Answer: \int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = \left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0

Then, replacing with point P and normal vector n:

36(x-5) + 20(y-0) + 45(z-0) = 0

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

z = \frac{180-36x-20y}{45}

When z = 0:

y = 9 + \frac{-9x}{5}

When z=0 and y=0:

x = 5

Then, triple integral is:

\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx

Calculating:

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx

\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0  {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx

\frac{1}{45} \int\limits^5_0  {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4}  } \, dx

\frac{1}{45} [30375-60750+118462.5-39150]

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx = 1087.5

<u>The volume of the tetrahedon is 1087.5 cubic units.</u>

3 0
3 years ago
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