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cricket20 [7]
4 years ago
5

Please help I’ve been stuck on this question and can’t seem to figure Out how to do It

Mathematics
2 answers:
Ipatiy [6.2K]4 years ago
7 0

Just divide the top on the right and go to calculate what the answer is

Ratling [72]4 years ago
5 0

Answer:

Just divide the numbers in the top right boxes.

tep-by-step explanation:

Please vote brainliest

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Which expression is the simplest form of −(2x 3 +x 2 )+3(x 3 −4x 2 )?
andreev551 [17]

Answer:

=x3−13x2

Step-by-step explanation:

8 0
3 years ago
What happens to the center of a figure when it is dilated about the origin?
alina1380 [7]
There are 2 answers for this in my opinion
1. nothing happens because its still in the origin and that won't change
2. depending on which way you turn it the points will change based on that

so if you know how to use a graph you should be good
7 0
4 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
3a.) Today, everything at a store is on sale. The store offers a 20% discount. 1 poin
Novay_Z [31]

Answer:

14.40

Step-by-step explanation:

100%-> 18

100-20=80

80%-> 14.40

6 0
3 years ago
B=ym+y for m   m=   help me solve thank u
Sveta_85 [38]
B = ym + y

B - y = ym

(B - y) / y = m
or
B/y - 1 = m


7 0
4 years ago
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