SWest

Here guest,

In order to find the sum of the coefficients, you need to simplify.

3(3x⁷+8x⁴-7) + 7(x⁵+7x²+5)

= 9x⁷+24x⁴-21+7x⁵+49x²+35

So the sum of the coefficients are:

9+24+7+49 = 89

So the sum of the coefficients is 89

Here you go guest:

if m² + 1 = 3m

we can simplify like this:

(m⁵+2m³-9m+27)/3m

= (m³(m²+2)-9m+27)/3m

= (m³(3m+1)-9m+27)/3m

= (3m⁴+(m³-9m)+27)/3m

= (3m⁴+m(m²-9)+27)/3m

= (3m⁴+m(3m+10)+27)/3m

= (3m⁴+3m²+10m+27)/3m

= (3m²(m²+1)+10m+27)/3m

= (3m²(3m)+10m+27)/3m

= (9m³+10m+27)/3m

= (m(9m²+10)+27)/3m

= (m(27m+1)+27)/3m

= (27m²+m+27)/3m

= (27(m²+1)+m)/3m

= (27(3m)+m)/3m

= (81m+m)/3m

=(82m)/(3m)

= 82/3

= 27 1/3

P.S. I probably made a mistake somewhere, but this is the general idea of how to do it!

Here you go guest.

If this is the graph of f(x)

f(-2) would mean the y-coordinate when x = 2

SO the answer is (2,-2) for the coordinate

SO the answer is 

-2

Thanks Wsai, here is another way to do it

Divisibility rule for 72 : Divisible by 8 and 9

\(\overline{a679b} \) = 10000a+6000+700+90+b

so we have two equations:

(790+b)/8 is an integer

a + 6+7+9+b is divisible by 9

so a=3 and b=2

Check

(790+2)/8 = 99 :)

(3 + 6  + 7 + 9 + 2)/9 = 3:)

Here you go guest:

|x-2| + |x-7| - |x-8| = 5

x=-4, x=22/3

If we want a quadratic equation to have 1 root, the discriminant must be 0.

so, b² - 4ac = 0

a=3

b=4

c = ?

4² - 4(3)(c) = 0

16-12c = 0

-12c=-16

c=16/12

c=4/3

In order for the domain to not work, the denominator.

x²-5x-4=0

\(x = {5 \pm \sqrt{25+16} \over 2}\)

A = (5+√41)/2

B = (5-√41/2)

A+B = (5+√41+5-√41)/2

A+B = 5

First, you need to find the solutions.

3x²+5x+7=0

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-5 \pm \sqrt{5^2-4*3*7} \over 2*3}\)

x = -(5-i√59)/6, and -(5+i√59)/6

We can rewrite as:

v = -(5+i√59)/6

 u = -(5-i√59)/6

u²+v² = (-(5-i√59)/6)²+ (-(5+i√59)/6)²

= -17/9

SWest

I'm assuming you want this simplified.

So 5x^2-3x-2 + x^2+2-5

= 6x^2-3x-2-3

= 6x²-3x-5

SWest