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2 years ago

Assist with this please

Mathematics

2 answers:

Natasha_Volkova [10]2 years ago

4 0

Answer:

The second line

Step-by-step explanation:

The answer to this equation is about 1.1, so the best line would be the second one

nikitadnepr [17]2 years ago

3 0

Answer:

iisbshdhdhhdhdhdhdhhdhdhhhdhhehehehhe

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Please help!! It’s timed!!

wariber [46]

Answer:

1/2

Step-by-step explanation:

8 0

3 years ago

3.452 × 100 =-------------

Lilit [14]

Answer:

345.2

Step-by-step explanation:

1: Move 3.452 to the end of the equation

(3.452 × 100) -> (3.452 × 100 = 3.452)

2: Since 100 has 2 zeros, move the decimal point 2 places to the right.

(3.452 × 100 = 3.452) -> (345.2)

Hope this helps!

7 0

3 years ago

Read 2 more answers

PlZ HELP WILL MARK YOU AS THE BRAINLIEST!

defon

Answer:

1. distance = sqrt( (7-7)^2+(2- -8)^2) = 10

2. check out desk (0,0 ) => distance = sqrt( (0- -9)^2+(0-0)^2) = 9

3. last corner ( -3, 4)

4. area = sqrt( (-10- -10)^2+(10-4)^2) x sqrt( (-3- -10)^2+(10-10)^2) = 6x7 =42

5. check desk (0,0), south direction = negative y axis => P_beginning (0,-20), P_end (0,-(20+25)) = (0,-45)

6. A(-2,-1) and B(4,-1) lie in y =-1. AB = sqrt( (-2- 4)^2+(-1- -1)^2) =6

=> area = 3.6x6 =21.6

=> peri = 2x(3.6+6) = 19.2

7. A(-5,4) and B(2,4), AB = sqrt( (-5- 2)^2+(4- -4)^2) = 7 => AB is base

=> p = peri = 7+ 8.3x2 = 23.6

=> area = sqrt[px(p-7)x(p-8.3)x(p-8.3)]

=sqrt[23.6x(23.6-7)x(23.6-8.3)x(23.6-8.3)] = 302.8

5 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

Given the parent function p(x) = cos x, which

GaryK [48]

Answer: left a units, down b units

Step-by-step explanation:

5 0

2 years ago

Assist with this please​

Assist with this please