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3 years ago

Can I get some help on these please

Mathematics

2 answers:

Gekata [30.6K]3 years ago

8 0

Answer:

1.2/5

2. 5/8

3. Proper Fraction

4. Mixed Fraction

6. 23/5

7. 19/5

8. 11 2/3

9. 50 4/5

Gelneren [198K]3 years ago

4 0

Answer:

1. $\frac{2}{5}$

2. $\frac{5}{8}$

3. Proper Fraction

4. Mixed Fraction

6. $\frac{23}{5}$

7. $\frac{19}{5}$

8. 11 $\frac{2}{3}$

9. 50 $\frac{4}{5}$

10. $\frac{1}{8}$

Sorry, didn't see the last question

Hope this helped!

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The sum of 2 numbers us 19 and their didfference is 55 what are the two numbers

damaskus [11]

Let one no. be x.
then another no. = (19-x).
x - (19 - x ) = 55
2 x - 19 = 55
x = 37
another no. = 19 - 37 = -18

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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

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

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myrzilka [38]

Answer:

Step-by-step explanation:

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Therefore, (x² - 10x + 25) will be the answer.

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For a perfect square of the given expression value of a should be 6.

x² + 2(a)x + 6² = x² + 2(6)x + 6²

= (x + 6)²

Therefore, x² + 12x + 36 will be the answer.

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To make this expression a perfect square,

a² = $(\frac{1}{4})^2$

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FrozenT [24]

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