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2 years ago

Write an equation for each translation of y = lxl .

Mathematics

2 answers:

Marizza181 [45]2 years ago

7 0

Answer:

the answer is y=|x-7|

Step-by-step explanation:

we know that the general translation of some equation y=f(x) to the left or right direction by 'a' units is given by y=f(x+a) ; where if 'a' is negative then the translation is in the "right direction" and if 'a' is positive then the translation is in the "left direction".

so in the given question we have the equation y=|x| for it to be translated 7 units in the right the resulting equation is y=|x-7|.

Alexxandr [17]2 years ago

5 0

Y=|x-7| because in order to go to the right you have to subtract the amount from x

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Jamal has run 7 miles from point a to point b of a race. the entire race from point a to point c is 26 miles. how many

Anon25 [30]

The number of more miles he needs to run to complete the race is 19 miles.

<h3 /><h3>How to find distance in miles?</h3>

He has run 7 miles from point a to point b of a race.

The entire race from point a to point c is 26 miles.

Therefore, the number of more miles he needs to run to complete the race is as follows:

let

x = the more miles he needs to run

Hence,

x = 26 - 7

x = 19 miles

Therefore, he needs to run 19 more miles to complete the race.

learn more on distance(miles) here: brainly.com/question/9950097

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1 year ago

Round 4358 to the nearest thousand

Hitman42 [59]

Answer:

4,000

Step-by-step explanation:

You go to the thousand number and see if the hundred number is higher than 5, if not then its stays the same.

5 0

3 years ago

Read 2 more answers

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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3 years ago

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liubo4ka [24]

Answer:

"I dont understand"

Step-by-step explanation:

What goes around comes around

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2 years ago

What is an equation in standard form of an ellipse centered at the origin with vertex (-6,0) and co-vertex (0,4)

qwelly [4]

$\frac{x^{2} }{36}$ + $\frac{ y^{2}}{16}$ = 1

This is because the standard form of an ellipse is

$\frac{ x^{2} }{ a^{2} }$ + $\frac{ y^{2} }{ b^{2} }$ = 1

where a is the vertex and b is the co-vertex. So when we stick their respective x and y values in and then square them, you're left with the answer above.

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3 years ago