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3 years ago

How many yards are equal to 240 feet

Mathematics

1 answer:

HACTEHA [7]3 years ago

8 0

80 yards equal 240 feet
You have to divide 240 by 3 since there are 3 feet in a yard

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I need help again sorry-

Margaret [11]

Answer:

f(x) = - 2(x - 3)² + 2

Step-by-step explanation:

Given

f(x) = - 2x² + 12x - 16 ( factor out - 2 from the first 2 terms )

= - 2(x² - 6x) - 16

To complete the square

add/subtract ( half the coefficient of the x- term)² to x² - 6x

f(x) = - 2(x² + 2(- 3)x + 9 - 9) - 16

= - 2(x - 3)² + 18 - 16

= - 2(x - 3)² + 2

4 0

3 years ago

6. John took Ms. Wisth out for sushi and is NOT a cheapskate. Their total sushi bill was $75.68. John

Anit [1.1K]

I think the answer would be $9.46 left as a tip for the waiter.

6 0

3 years ago

Read 2 more answers

Which measure of center best represents this set of data? Number of television sets in the home: 4, 4, 2, 4, 4, 1, 3, 4, 2, 4 *

nataly862011 [7]

Answer:

Mean

Step-by-step explanation:

Given

$Data: 4, 4, 2, 4, 4, 1, 3, 4, 2, 4$

Required

Best measure of center

The mean is considered the best measure of center when there are no outliers in the given data.

Outlier is when a particular data set is far away from other datasets.

For instance: 20 is an outlier in the following data sets: $4, 4, 2, 20, 4, 4, 1, 3, 4, 2, 4$

Also, the given data are not open-ended.

An open-ended data is as follows: $1, 2, 3, 4, 5\ or\ more$

Since we do not have any of the above conditions, then the mean is the best center to use.

7 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

Question 3 of 10

Aleks [24]

C=+3 since your adding three more onto four

4 0

3 years ago