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The graph is shifted left 6 units.

When the expression  $(2^1)(2^2)(2^3)\cdots (2^{99})(2^{100})$    is written as an integer,

what is the product of the tens digit and the ones digit?

1+2+3+.....100 = 100/2(1+100)= 50*101 = 5050

so this is 

   $2^{5050}=(2^{25*202})\\ =((2^{25})^{202})\\ \text{now only considering the last two digits}\\ \equiv (32)^{202}\\ \equiv (32)^{101*2}\\ \equiv (32^2)^{101}\\ \equiv (24)^{101}\\ \equiv 24*(24)^{100}\\ \equiv 24*((24)^{5})^{20}\\ \equiv 24*(24)^{5*4}\\ \equiv 24*24^4\\ \equiv 24$

This is probably not the best way to do it though :/

2^1 x 2^2 x2^3 x.........x 2^99 X 2^100 =2^(1+2+3+.......+99+100) =2^(100 x 101) / 2 =2^5050.

Any power of 2 ending in: 10, 30, 50, 70, 90 will have the last 2 digits as =.....24.

(-1,5):   a(-1)²+b(-1)+c = 0 or a - b + c = 0

(0,5):    a(0)²+b(0)+c = 5 or c = 5

(5,0):   a(5)²+b(5)+c = 0 or 25a + 5b +c = 0

So: 

a - b + 5 = 0.        (1)

25a + 5b + 5 = 0. (2) 

Multiply (1) by 5 and add to (2):

30a + 30 = 0.  So a = -1 

Hence, From (1): 

-1 - b + 5 = 0. So b = 4

a = -1, b = 4, c = 5

I’ll leave you to calculate 100a + 10b + c

For the SECOND one you will also have 12 waveforms in 2 pi ...but it is a SIN wave

it will have an amplitude of 3  (6 units from peak to low point)   and it will be shifted DOWN by 5 units....

Cool?

Here ya go.....

OK.... due the the 12x  it will have 12 complete waveforms in a x distance of 2pi

The amplitude will be  '7'       or 14 from the top to the bottom of the wave

And finally the cos wave will be shifted UP by 1

I will post a pic in a sec

It appears as if $\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}$ is the best combination.

By the way, $\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}\approx1.99802$

1.

Simplify the following:

6/(2 sqrt(3) - 3)

Multiply numerator and denominator of 6/(2 sqrt(3) - 3) by the conjugate of the denominator.

Multiply numerator and denominator of 6/(2 sqrt(3) - 3) by 2 sqrt(3) + 3:

(6 (2 sqrt(3) + 3))/((2 sqrt(3) - 3) (2 sqrt(3) + 3))

Multiply 2 sqrt(3) - 3 and 3 + 2 sqrt(3) together using FOIL.

(2 sqrt(3) - 3) (2 sqrt(3) + 3) = -3×3 - 3×2 sqrt(3) + 2 sqrt(3)×3 + 2 sqrt(3)×2 sqrt(3) = -9 - 6 sqrt(3) + 6 sqrt(3) + 12 = 3:

(6 (2 sqrt(3) + 3))/3

In (6 (2 sqrt(3) + 3))/3, divide 6 in the numerator by 3 in the denominator.

6/3 = (3×2)/3 = 2:

2 (2 sqrt(3) + 3)

2.

Simplify the following:

1/(sqrt(2) - 1) + 1/(sqrt(2) + 1)

Multiply numerator and denominator of 1/(sqrt(2) - 1) by the conjugate of the denominator.

Multiply numerator and denominator of 1/(sqrt(2) - 1) by sqrt(2) + 1:

(sqrt(2) + 1)/((sqrt(2) - 1) (sqrt(2) + 1)) + 1/(sqrt(2) + 1)

Multiply sqrt(2) - 1 and 1 + sqrt(2) together using FOIL.

(sqrt(2) - 1) (sqrt(2) + 1) = -1 - sqrt(2) + sqrt(2)×1 + sqrt(2) sqrt(2) = -1 - sqrt(2) + sqrt(2) + 2 = 1:

(sqrt(2) + 1)/1 + 1/(sqrt(2) + 1)

Look for one in the denominator of sqrt(2) + 1.

(sqrt(2) + 1)/1 = sqrt(2) + 1:

sqrt(2) + 1 + 1/(sqrt(2) + 1)

Multiply numerator and denominator of 1/(sqrt(2) + 1) by the conjugate of the denominator.

Multiply numerator and denominator of 1/(sqrt(2) + 1) by sqrt(2) - 1:

1 + sqrt(2) + (sqrt(2) - 1)/((sqrt(2) + 1) (sqrt(2) - 1))

Multiply 1 + sqrt(2) and sqrt(2) - 1 together using FOIL.

(sqrt(2) + 1) (sqrt(2) - 1) = -1 + 1 sqrt(2) - sqrt(2) + sqrt(2) sqrt(2) = -1 + sqrt(2) - sqrt(2) + 2 = 1:

1 + sqrt(2) + (sqrt(2) - 1)/1

Look for one in the denominator of sqrt(2) - 1.

(sqrt(2) - 1)/1 = sqrt(2) - 1:

1 + sqrt(2) + sqrt(2) - 1

Evaluate 1 + sqrt(2) - 1 + sqrt(2).

1 + sqrt(2) - 1 + sqrt(2) = 2 sqrt(2):

2 sqrt(2)

Just to clarify. There are 8 variables, half are divided by the other half, and they are all different single digit numbers. Together, in the previously mention format, they need to be less than 2, but as big as possible.

Simplify the following:

(-2 i + 1) (-i + 1) (i + 1) (2 i + 1)

(1 - 2 i) (1 - i) = 1×1 + 1 (-i) + -2 i×1 + -2 i (-i) = 1 - i - 2 i - 2 = -1 - 3 i:

-3 i - 1 (1 + i) (2 i + 1)

Factor -1 from -3 i - 1:

-(3 i + 1) (1 + i) (2 i + 1)

(1 + 3 i) (1 + i) = 1×1 + 1 i + 3 i×1 + 3 i×i = 1 + i + 3 i - 3 = -2 + 4 i:

-4 i - 2 (2 i + 1)

Factor 2 out of 4 i - 2 giving 2 (2 i - 1):

-2 (2 i - 1) (2 i + 1)

(-1 + 2 i) (1 + 2 i) = -1 - 2 i + 2 i×1 + 2 i×2 i = -1 - 2 i + 2 i - 4 = -5:

-2×-5

-2 (-5) = 10

thanks

7x-(x-5) ≥ 14     simplify

7x - x + 5 ≥  14     combine like terms

6x + 5 ≥ 14     subtract  5 from both sides

6x ≥ 9        divide both sides by 6

x ≥  9/6  ⇒   x  ≥ 3/2

Right back at ya Foxy. 😃

OK.....I'm gonna' answer this.....you just can't trust a procrastinator like that darn hectictar....(HEHEHEHE!!)

(10, 2)  and (8, 7)

Slope

(1) Subtract the y coordinates in some order  ⇒ 7 - 2   =  5        ....    (1)

(2) Subtract the x coordinates in the SAME order ⇒   8 - 10 =   - 2   ...    (2)

Put   (1)  over (2)

5 / -2     =   -5 / 2        and that's the slope

Oh no!!! I said I was gonna do better about always waiting till the last minute!!! Haha

Oops...you're late, hectictar....it's already 12:07 !!!  LOL!!!!!!!

See???.....you already got a point....!!!

451^2  -  137^2  =

(451 + 137)(451 - 137)

(588)(314)

(294)(2)(157)(2)

So........the largest prime factor is  157

HURRY!!! I HAVE UNTILL 12:06

1.

f(x) + g(x)

                                                   Since  f(x)  =  -5x - 4, we can substitute  -5x - 4  in for  f(x)

f(x) + g(x)  =  (-5x - 4) + g(x)

                                                   Since  g(x)  =  6x - 7, we can substitute  6x - 7  in for  g(x)

f(x) + g(x)  =  (-5x - 4) + (6x - 7)

f(x) + g(x)  =  -5x - 4 + 6x - 7

                                                   Combine like terms.

f(x) + g(x)  =  x - 11

2.

f(x) - g(x)

                                                   Since  f(x)  =  3x + 2, we can substitute  3x + 2  in for  f(x)

f(x) - g(x)  =  (3x + 2) - g(x)

                                                   Since  g(x)  =  6x - 7, we can substitute  6x - 7  in for  g(x)

f(x) - g(x)  =  (3x + 2) - (6x - 7)

                                                   Distribute the  -1  to the terms in parenthesees.

f(x) - g(x)  =  3x + 2 - 6x + 7

                                                   Combine like terms.

f(x) - g(x)  =  -3x + 9

4.

$\frac{f}{g}\,=\,\frac{f(x)}{g(x)} \\~\\ \frac{f}{g}\,=\,\frac{x^2-16}{x+4}$

Before simplifying, let's get the domain.  x + 4  is in a denominator, so...

x + 4 ≠ 0

x  ≠  -4     The domain is all real numbers with the exception that  x ≠ -4

$\frac{f}{g}\,=\,\frac{x^2-16}{x+4}$

                                   Factor the numerator as a difference of squares.
$\frac{f}{g}\,=\,\frac{(x-4)(x+4)}{x+4}$

                                   Reduce the fraction by  (x + 4) .

$\frac{f}{g}\,=\,\frac{(x-4)}{1} \\~\\ \frac{f}{g}\,=\,x-4$