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polet [3.4K]
3 years ago
10

A total of 366 tickets were sold for the school day they were pay they were either adults tickets or students tickets.the number

of student tickets.the number of student tickets sold was two times the number of adults tickets sold.how many adult tickets were sold ?
Mathematics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

122

Step-by-step explanation:

366 divided by 3


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The sum of 237 and six times the opposite of a number is 405. What is the number?
Bond [772]

Answer:

(A) x=-28

Step-by-step explanation:

237 -6x = 405           [note that 6*(-x) is (-6x)]

  -6x = 168           [subtract 237 from both sides]

   x = -28             [divide both sides by (-6)]

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S_A_V [24]
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3 years ago
Use a calculator to find the value of these data. Round the value to three
insens350 [35]

Answer:

-1, -1.1, 0.9, 1

Step-by-step explanation:

6 0
3 years ago
The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each
Fynjy0 [20]

Answer:

\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Step-by-step explanation:

Given equation of ellipsoids,

u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}

The vector normal to the given equation of ellipsoid will be given by

\vec{n}\ =\textrm{gradient of u}

            =\bigtriangledown u

           

=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})

           

=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}

           

=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}

Hence, the unit normal vector can be given by,

\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}

             =\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}

             

=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Hence, the unit vector normal to each point of the given ellipsoid surface is

\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

3 0
3 years ago
To graph the line y = 3x -2, I should *
ipn [44]

Answer:

The answer is D because you should starts at -2 on the y-axis and moves up 3 and right 1.

4 0
4 years ago
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