All Answers 358183 Answers

\(z \in \{z|\ (-1) < z < 1 \}\)

  !

It may show on your computer, but no one else can see it.

This is what we see:

4) In a coordinate system, the magnitude of the x component of a vector
and , the angle between the vector and x-axis, are known. The
magnitude of the vector equals the x component

a. divided by the cosine of .
b. divided by the sine of .
c. multiplied by the cosine of .
d. multiplied by the sine of .

Whats missing? That is the entire question on the summer assignment.

No one can answer your question. There are too many blanks in your post, Blank.

Consider a circle, radius 1 , with its centre on the line OB .

Call the centre b' and start with b' at O.

Move b' away from O.

The maximum distance Ob' will happen when the line OA is tangential to the circle.

When that happens the line from OA to OB will be at right angles to OA.

Ob' will then be the hypotenuse of a 30 deg right-angled triangle.

In that triangle, 1/Ob' = sin(30 deg) =1/2.

So 0b' = 2.

roger

7 and 8 are relatively prime so the solutions will be separated by their product, i.e. 56.

35 + 2 x 56 = 147

so adding 53 gets us exactly 200, so it must be the next one

35+3x56 = 203

is the first solution that meets the requirements.

The original number of coins was 203.

If there are no restrictions, then:

[6 + 3 - 1] C [3 - 1] =8C2 =28 different ways.

Because, if you fator 1000!, you will see that 7^164 is one of its factors.

The largest n = 164

My teacher posted the answer as (A) but I dont understand how he got that answer. Please help.

Find the set of all z such that neither 2+z nor 2-z is in the interval (-1,1]. Express your answer as an interval or as a union of intervals.

z \in \{z|\  (-1) < z < 1 \}

  !

LaTeX does not work.

Rom: The question says " You have more than 200 coins AFTER the  8th bag of 53"

"The sum of x and its reciprocal" is equal to \(x+\frac1x\), which simplifies to \(\frac{x^2+1}{x}\).  We know that from the equation, the numerator is equal to 7x.  7x/x is simply equal to 7.

\(\text{Well let's turn this into a couple equations if nothing else}\\ N = 7j \\ N = 0 \pmod{7}\\ N+53 = 8k\\ N+56-3=8k\\ N-3=8(k-7)\\ N=3 \pmod{8}\)

\(\text{So we have the system}\\ N=0 \pmod{7}\\ N=3 \pmod{8}\)

\(\text{Just by inspection $N=35$}\)

For more detailed explanation of the "Chinese Remainder Theorem + Modular Multiplicative Inverse", see this link here: https://web2.0calc.com/questions/congruences

Yes, thank you!

I had just got it right when I saw yours, so I can verify.

I reemember this one from several days ago. 

N = 98            divisible by 7² 

N+1 = 99        divisible by 3²  

N+2 = 100      divisible by 5² and by 2²       

.

Just count them!

(1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14, 15, 16, 20, 21, 22, 23, 24, 25, 26, 30, 31, 32, 33, 34, 35, 36, 40, 41, 42, 43, 44, 45, 46, 50, 51, 52, 53, 54, 55, 56, 60, 61, 62, 63, 64, 65, 66, 100, 101, 102, 103, 104, 105, 106, 110, 111, 112, 113, 114, 115, 116, 120, 121, 122, 123, 124, 125, 126, 130, 131, 132, 133, 134, 135, 136, 140, 141, 142, 143, 144, 145, 146, 150, 151, 152, 153, 154, 155, 156, 160, 161, 162, 163, 164, 165, 166, 200, 201, 202, 203, 204, 205, 206, 210, 211, 212, 213, 214, 215, 216, 220, 221, 222, 223, 224, 225, 226, 230, 231, 232, 233, 234, 235, 236, 240, 241, 242, 243, 244, 245, 246, 250, 251, 252, 253, 254, 255, 256, 260, 261, 262, 263, 264, 265, 266, 300, 301, 302, 303, 304, 305, 306, 310, 311, 312, 313, 314, 315, 316, 320, 321, 322, 323, 324, 325, 326, 330, 331, 332, 333, 334, 335, 336, 340, 341, 342, 343)

Thank you very much!!!   

After a slight re-arrangement of your congruences, you have the following congruences, which are the same.
x mod 3^3 = 1
x mod 5^3 = 4
x mod 7^3 = 18 
Using "Chinese remainder theorem" + "Modular multiplicative inverse", which are inculded in this short computer code, the value of x=
1157625 m +  506629, where m=0, 1, 2, 3......etc.
i=0;j=0;m=0;t=0;a=(27, 125, 343);r= (1, 4, 18);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return.
So, the smallest value for x =506,629
506,629 mod 105 =2^2 - the remainder.

165 degrees   magnitude   270    firnd the x and y components

x = 270 cos 165 =  - 260.8

y = 270 sin 165 = 69.9  

203 degrees magnitude 130   find the x and y components

x = 130 cos 203  =  -119.7    (edited)

y=  130 sin 203 =  -50.8        (edited)

Add the x components and the y components

x =- 260.8+-119.7 = -380.5

y = 69.9 - 50.8 = 19.1

d = sqrt (x^2 + y^2) = 381 km  from origin   bearing = arctan (y/x) = 177.1 degrees

xxxxxxxxxxxxxxx

I don't know, I just randomly made up some time.

This is not a off topic post, i just accidently clicked it.