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3 years ago

A total of 19,035 people skied at the resort during the 5 days

Mathematics

2 answers:

maw [93]3 years ago

7 0

Answer:

3805

Step-by-step explanation:

To find the average over a period of time we divide 19,035 by the amount of time skied. Which is 5 days.

19035/5 is 3805

LiRa [457]3 years ago

6 0

3805 is the correct answer to your question

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Is square root of 7 a rational number?? Why or why not.

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3 years ago

Find two rational numbers between 5/6 and 83/ 100 . Explain how you know (without using decimals)

kolbaska11 [484]

Answer:

998/1200 and 997/1200

Step-by-step explanation:

It’s hard to compare 5/6 and 83/100 in their current states. Let’s find a common denominator. Take 600. We know that this will work because both 6 and 100 go into it. Multiplying 5 by 100 and 83 by 6 to find the common denominator, we can compare them at 498/600 and 500/600. However, there is only one number (499/600) between them. Thus, we will multiply each number by 2/2, which is 1. Our numbers are then 996/1200 and 1000/1200. Two numbers between them are then 997/1200 and 998/1200

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3 years ago

I’ve been stuck on question 1 for a while, can you guys help?

MakcuM [25]

Answer:

$h=20$ **

**The picture is not accurate for this problem.

Step-by-step explanation:

I would use Pythagorean Theorem to setup two equations since there are two triangles with no angle information.

Let $BC=x+(99-x)$ where $x$ is equal to the first partition of $BC$ (reading from left to right) and $(99-x)$ is equal to the second partition of $BC$ (reading from left to right).

We have the following system to solve:

$20^2=h^2+x^2$

$101^2=h^2+(99-x)^2$

I will use elimination to first solve for $x$ .

Subtract the equations:

$20^2-101^2=x^2-(99-x)^2$

Factor both sides using $a^2-b^2=(a-b)(a+b)$ :

$(20-101)(20+101)=(x-(99-x))(x+(99-x))$

Simplify inside the $($ $)$ .

$(-81)(121)=(2x-99)(99)$

Divide both sides by 9:

$(-9)(121)=(2x-99)(11)$

Divide both sides by 11:

$(-9)(11)=(2x-99)(1)$

Simplify both sides:

$-99=2x-99$

Add 99 on both sides:

$0=2x$

Divide both sides by 2:

$x=0$

Now go to either equation we had in the beginning to find $h$ .

$20^2=h^2+x^2$ with $x=0$

$20^2=h^2$

$20=h$

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3 years ago

Which of the following could be points on the unit circle: A. (4/3, 4/5) B. (5/13, 12/13)

IRINA_888 [86]

Answer:

Option (2) and (4) are correct.

$(\frac{5}{13},\frac{12}{13} )$ and $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ are points on the unit circle.

Step-by-step explanation:

Given : Some points of circles.

We have to choose which points could be points on the unit circle.

We know, the equation of circle is

$x^2+y^2=r^2$ ..........(1)

where r is radius of circle.

We check each point for x and y values on the equation of circle and see which point gives radius = 1

Thus,

1) $(\frac{4}{3},\frac{4}{5} )$

Put in LHS of (1) , we have,

$(\frac{4}{3})^2+(\frac{4}{5})^2$

Simplify, we have,

$=\frac{16}{9}+\frac{16}{25}$

$=\frac{544}{225}\neq 1$

Thus, $(\frac{4}{3},\frac{4}{5} )$ is not a point on the unit circle.

2) $(\frac{5}{13},\frac{12}{13} )$

Put in LHS of (1) , we have,

$(\frac{5}{13})^2+(\frac{12}{13})^2$

Simplify, we have,

$=\frac{25}{169}+\frac{144}{169}$

$=\frac{169}{169}= 1$

Thus, $(\frac{5}{13},\frac{12}{13} )$ is a point on the unit circle.

3) $(\frac{1}{3},\frac{2}{3} )$

Put in LHS of (1) , we have,

$(\frac{1}{3})^2+(\frac{2}{3})^2$

Simplify, we have,

$=\frac{1}{9}+\frac{4}{9}$

$=\frac{5}{9}\neq 1$

Thus, $(\frac{1}{3},\frac{2}{3} )$ is not a point on the unit circle.

4) $(\frac{6}{7},\frac{\sqrt{13}}{7} )$

Put in LHS of (1) , we have,

$(\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2$

Simplify, we have,

$=\frac{36}{49}+\frac{13}{49}$

$=\frac{49}{49}= 1$

Thus, $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ is a point on the unit circle.

Thus, $(\frac{5}{13},\frac{12}{13} )$ and $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ are points on the unit circle.

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3 years ago

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