a.

By Fermat's little theorem, we have


5 and 7 are both prime, so
and
. By Euler's theorem, we get


Now we can use the Chinese remainder theorem to solve for
. Start with

- Taken mod 5, the second term vanishes and
. Multiply by the inverse of 4 mod 5 (4), then by 2.

- Taken mod 7, the first term vanishes and
. Multiply by the inverse of 2 mod 7 (4), then by 6.


b.

We have
, so by Euler's theorem,

Now, raising both sides of the original congruence to the power of 6 gives

Then multiplying both sides by
gives

so that
is the inverse of 25 mod 64. To find this inverse, solve for
in
. Using the Euclidean algorithm, we have
64 = 2*25 + 14
25 = 1*14 + 11
14 = 1*11 + 3
11 = 3*3 + 2
3 = 1*2 + 1
=> 1 = 9*64 - 23*25
so that
.
So we know

Squaring both sides of this gives

and multiplying both sides by
tells us

Use the Euclidean algorithm to solve for
.
64 = 3*17 + 13
17 = 1*13 + 4
13 = 3*4 + 1
=> 1 = 4*64 - 15*17
so that
, and so 
<em><u>An inequality that shows the distance Johnathan could of ran any day this week is:</u></em>

<em><u>Solution:</u></em>
Let "x" be the distance Johnathan can run any day of this week
Given that,
Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles
Therefore,
Number of days ran = 5
The most he ran in 1 day = 3.5 miles
Thus, the maximum distance he ran in a week is given as:

The maximum distance he ran in a week is 17.5 miles
If we let x be the distance he can run any day of this week then, we get a inequality as:

If we let y be the total distance he can travel in a week then, we may express it as,

I cant really show you how to graph this but i can try to explain it. you would make a 4 quadrant graph and do rise over run.
1)start by making the graph
2) on the line of the graph marked Y is where you would put -4
3) From -4, you would go up (rise) by 2 and go over to the right 1 where you end is the slop
Answer:
x = ± 2i
Step-by-step explanation:
The equation has no real roots, gut has complex roots
Given
x² = - 4 ( take the square root of both sides )
x = ± 
= ± 
[ Note that
= i ]
= ±
×
= ± 2i