Alan

The general expression for a quadratic function can be written as  y = a*x² + b*x + c where a, b and c are constants. 

You can find a, b and c by using the points on the graph

2 = a*(-2)² + b*(-2) + c     or  2 = 4a -2b + c

1 = a*0² + b*0 + c             or  1 = c

-2.5 = a*1² + b*1 + c         or -2.5 = a + b + c

You now have 3 equations for the 3 unknowns a, b and c.  Can you take it from here?

Here are the steps:

\(5\sqrt{18}-2\sqrt{2}\rightarrow5\sqrt{2\times9}-2\sqrt{2}\rightarrow \sqrt{2}(5\sqrt{9}-2)\rightarrow \sqrt{2}(5\times3-2)\rightarrow13\sqrt{2}\)

Assuming this is (5x - 3)/6 = x + 2 then multiply both sides by 6

5x - 3 = 6x + 12

subtract 5x from both sides

-3 = x + 12

subtract 12 from both sides

-15 = x

or just x = -15

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Use the fact that \((a\times b)^m = a^m\times b^m\)

and the fact that \((a^m)^n=a^{m*n}\)

to get \(-(2x^2)^5\rightarrow -(2^5\times (x^2)^5) \rightarrow -(32\times x^{10})\)

The resistive force is (coefficient of friction)*mass*(gravitational acceleration), so the net force (F) minus this resistive force must equal mass*acceleration; or:

F - 0.85*3*9.81 = 3*0.15

F = 3*0.15 + 0.85*3*9.81

F = 25.47 Newtons

Try:  y = 2x + 18

x = 1  y = 20

x = 2  y = 22

x = 3  y = 24

...etc.

It can be factored as follows:

\(7k^2+9k \rightarrow k(7k+9)\)

Use of brackets would clarify the question!

\(4^{\frac{1}{2}+\frac{a}{2}}\\\\ =4^{\frac{1}{2}(1+a)}\\\\ =2^{1+a}\\\\ =2\times2^a\)

Not sure that my answer is correct as my states wouldn't consist of just 1s and 0s.

Does this help:

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