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2 years ago

Find the circumference of the circle with the given area A. Use 3.14 for TI

1 answer:

4 0

Area = r^2 * 3.14
87.9/3.14 = around 28
Square root of 28 = 5.3
The radius is 5.3
Circumference: diameter * 3.14
5.3 + 5.3 = 10.6
10.6 * 3.14 = 33.3
The circumference is 33.3

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If parker uses online banking and average 20 transactions a month, which bank should he choose

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Each month that will give them 20$ for

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2 years ago

Simplify the expression-2.3 + (-5.7). On the test when maureen simplified the expression she got -3.4. What mistake did maureen

Hoochie [10]

Answer:

see explanation

Step-by-step explanation:

- 2.3 + (- 5.7)

reminder that + (- ) = -

To obtain - 3.4 it is likely that she added 2.3 to - 5.7

The solution is

- 2.3 - 5.7 = - 8

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2 years ago

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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

2 years ago

A horse-drawn carriage tour company has found that the number of people that take their tour depends on the price charged per cu

Mars2501 [29]

Answer:

The horse-drawn carriage tour company can expect to take in $6960 when the charge per customer is $60.

Step-by-step explanation:

p(2) = 120 -2·2 = 116 . . . . . expected number of customers per day

c(2) = 50 +5·2 = 60 . . . . . . charge per customer

Then ...

(p·c)(2) = p(2)·c(2) = 116·60 = 6960 . . . . revenue for the day

6 0

2 years ago

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Skylar invested $8,900 in an account paying an interest rate of 2.6% compounded

mamaluj [8]

Answer:

11847

Step-by-step explanation:

8900e ^0.026(11)

8900e^0.286

11846.7228518

answer 11847

4 0

3 years ago