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Since the line passes through (-8, -9), the line x = -8   would pass through this point.....and this line would also be parallel to x = 5

So...the line is x = -8

What is the smallest positive integer n such that n has remainder 1 when divided by 3, n has remainder 1 when divided by 4, and n has remainder 4 when divided by 5?

n = 60 m+49, m is any positive integer.

n = 60.1 +49 =109 is the smallest positive integer that satisfies all above conditions.

Note that :

9^1 mod 7 = 2

9^2 mod 7 = 4

9^3 mod 7 = 1

In general, for any integer n ≥ 0,

9^(1 + 3n) mod 7 = 2

9^(2 + 3n) mod 7 = 4

9^(3 + 3n)mod 7 = 1    

And 1995 = 3 + 3(664)

So.....9^1995 mod 7  = 1

1) Find the value “c” that shows the mean value theorem for derivatives holds for f(x) =2x^2 on the interval [−3,1] 

The slope of the line drawn between f(-3)  and f(1) =     [ 2 - 18] / [ 1 - (-3) ] =   -16/ 4 = -4

And the derivative of 2x^2  =  4x

So.....we're looking for the x value where the tangent line to the curve at that point has the slope of -4

4x = -4     divide through by 4

x = -1

And    -3 < -1 < 1       so the Theorem holds

2) Find the value “c” that shows the mean value theorem for integrals holds for 𝑓f(x) =2x^2 on the interval [1,6] 

We are trying to find "c"  such that

6

∫   2x^2 dx =   (2)c^2 * (6 -1)

1

 (2/3)[6^3 - 1^3]  = 10c^2

(2/3) (215)  = 10c^2

215/15 = c^2

43/3  = c^2   take the positive root

√[43/3]  = c ≈  3.7859

And this number is in the interval [1,6]

Awesome thanks heaps - have passed onto my friend

Type "(pi)^3" on this calculator and it will show \(\mathtt{\pi^3}\)

Then you press Enter on your keyboard and enjoy the answer given by this calc!! :D

What is 2 to the fith power?

\(2^5=2\times 2\times 2\times 2\times 2= 32 \)

\(\)

2^5 = 32 :D

This is question 3.

1^3+2^3+3^3+...+100^3 mod 6

= (1+2+3+4+5+6+7+8+........100)^2 mod 6

= 5050^2 mod 6

= 5050 mod 6

= 4.

1. Find the second smallest positive integer that gives a remainder of 2 when divided by 3 and gives a remainder of 3 when divided by 7.

We have that

N = 3a + 2

N = 7b + 3

Subtracting these equations we have that

3a - 7b - 1  = 0

3a - 7b =  1

And the second smallest positive integer results when a = 12 and b = 5

So.....the second smallest integer is 38.....just as Max said !!!!

It's for finding the surface area of a cylinder with base radius r and height h.

CPhill means you are correct

x= [0, 9].

#1 is correct.

#2 is wrong.

#2) g(f(-6))=g(-21)=-21 -2 = -23.

#3 is wrong.

#3) f(f(7))=f(31)=4(31) + 3 = 127 <- you forget to add that 3.

#4 is wrong but not completely wrong, you didn't simplify.

g(f(x)) = g(4x+3) = 4x+3 - 2 = 4x + 1.

1) We first have a list of what gives a remainder 2 when divided by 3 and we will have a list of what gives a remainder of 3 when divided by 7.

Gives a remainder 2 when divided by 3:

2,5,8,11,14,17,20,.......

Gives a remainder 3 when divided by 7:
3,10,17,......

Then we add 17 to the LCM of 3 and 7:

17 + 21 = 38

Answer : 38