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How many ordered pairs (x,y) satisfy BOTH conditions below?

 Condition I: x=1 or y=0 or y=2

 Condition II: x=0 or x=2 or x=1

Since all points where x=1 are included in both sets, there is an infinite number of points that satisfy both conditions.

ex  (1,-10)(1,-3)!1,0), 1,100) etc etc etc

The actual points that satisfy both contitions is all points where x=1 plus (0,0), (0,2),(2,0),(2,2)

Suppermanaccz,

It is good that you have stated that you do not understand.

BUT

You have not even said which answer you do not understand.  There are 2 different question!

You need to say WHAT question and also try to say where you get lost.

If you do not understand the first line then say so.

If you do understand the first line then tell GYanggg what you do not understand!

2 - 4 - 4

4 - 3 - 3

a. Assume that the bacteria are distributed uniformly over the​ planet's surface. How deep would the bacterial layer​ be.

1)

1/log_2(N) + 1/log_3(N) + 1/log_4(N) +..............+1/log_100(N)=1/log(100!*N)

Factor out N, then:

The above can written as:

1 / log[N * (1 x 2 x 3 x 4 x 5.................100)] = 1/ log[N*100!]

What are some of the less beautiful realms of mathematics?

huh?

you lost me

1.

From the definition of logarithm, we obtain:

\(\frac{1}{log_b{a}}=\log_a{b}\)

The equation can therefore be written in the form:

\(\log_n2+\log_n3+\cdots\log_n100+=\log_n(2\cdot3\cdots100)\)

from which the desired conclusion immediately follows. 

2. 

https://www.desmos.com/calculator/vp8etonnna 

First, if \(\sin{x}=\log{x},\)  then \(x\le10\) (inasmuch as \(\sin{x}\le1\). 

Since \(2\cdot2\pi>10\), the interval on the x-axis between x = 0 and x = 10 contains one complete period of the sine curve plug part of a second period. The graph of \(\log{x} \) intersects the first wave of the sine curve at precisely one point. Futher, since \(2\Pi+\frac{\Pi}{2}<10\), than at the point \(x=\frac{5\Pi}{2}\), we have \(\sin{x=1}>\log{x}\), which means that the graph of log x intersects the first half of the second positive wave of sin x. Since, at \(x=10, \log{x=1}>\sin{x}, \) the graph intersect this second wave another time. Therefore we conclude that the equation \(\sin{x}=\log{x},\) has exactly three roots.

Wow! I haven't done these in SO long! Thanks for bringing me back to the more beautiful realm of mathematics. 

I hope this helped,

Gavin

Can someone please help me?

THis is due tmr and I have no clue

I don't understand your question, can you repost in proper form?

Thanks CPhill!

THANK YOU SOOO MUCH!!!

The midpoint of the line connecting these two points will lie on the line we seek...and this is

[ (1 + 9 ) / 2, ( 1 + 5) / 2 ]  =  [ 5, 3 ]

And the slope of this line connecting the given points is  [ 5 - 1 ] / [ [ 9 - 1]  = 4/8  =  1/2

And the line we seek is perpendicular to this line...so its slope is  - 2

And the equation of this line  is

y = -2 (x - 5) + 3

y = - 2x  + 10 + 3

y  = - 2x + 13

So...m  +  b  =   -2 + 13   =  11 

Here's a pic :  https://www.desmos.com/calculator/jdelhysrdy

Thanks so much, tertre!

Let's say the x is the miles per hour for car A, and x+10 is the number of miles for car B(says in the problem/question)

Since Car A travels 120 miles when Car B travels 150 miles, we can do:\(\frac{120}{x}=\frac{150}{x+10}\).

Cross multiplying, we have 120(x+10)=150x- 120x+1200=150x

We get 30x=1200-x=40

Thus, Car A travels 40 mph, and Car B travels 50 mph.

Thank you both so much!

Let the distance from Post Office to Friend's house =D

7/8 + 5/6 + D =miles - Total distance from Georgia's house to a Friend's house.

42 minutes / 18 minutes per mile =2 1/3 miles - toal distance from her house to a friend's house. So, we have:

7/8 + 5/6 + D = 2 1/3, solve for D

[21 + 20] / 24 + D = 2 1/3

41/24 + D = 2 1/3                     subtract 41/24 from both sides

D =2 1/3 - 41/24

D =7/3 - 41/24

D =[56 -41] / 24

D = 15 / 24

D = 5/8 of a mile - distance between the Post Office and her Friend's house.

Thank you so much!

The answer was right BTW

this is very readable, I don’t understand what you mean

another user already answer, which means it is readable

Rollingblade, PLEASE make sure that your questions are readable, this is unbearable. You can't expect this forum to help you if you can't write your question instead of copying and pasting it.

Hey Rollingblade!

When N is divided by 10, the remainder is a, 

When N is divided by 13, the remainder is b.

From these two statements, we can setup the modular arithmetic systems:  

\(\begin{align*} x &\equiv 1 \pmod{10}, \\ x &\equiv 0 \pmod{13}. \end{align*}\)  

The solution is: \(X\equiv91\pmod{130}\)

Then we solve:

\(\begin{align*} y &\equiv 0 \pmod{10}, \\ y &\equiv 1 \pmod{13}. \end{align*}\) 

The solution is: \(y\equiv40\pmod{130}\)

Therefore, the answer is \(N\equiv91a+40b\pmod{130}\)

I hope this helped,

Gavin