Two points determine a line. In slope intercept form we have:
y = -2x + b
So:
-12 = -2(3) + b and [first equation]
k = -2(6) + b [second equation]
We have a system of simultaneous equations with two unknowns. The first is trivially easy to solve:
-12 = -6 + b [Evaluate -2(3)]
-6 = b [Add +6 to each side]
So we substitute this into the second equation:
k = -2(6) + -6 = -18
We could double check our work knowing that m=Δy/Δx
We expect m to be -2. Does:
-2 = (-18 - -12)/(6 - 3) = -6/3 = -2?
;)
sec ( t ) cos ( t ) = (1/cos (t)) *cos(t) = 1
Hello from MrBillDoesMath!
Answer:
x = 30
Discussion:
Note: I ignored the * in f*(x) as i don't know what it means. (If this is not what you intended, please indicate what * means)
f(x) = 2 sqrt(x-5) + 4
f(x) = 14
=>
14 = 2 sqrt(x-5) + 4 => subtract 4 from both sides
14 -4 = 2 sqrt(x-5) => divide both sides by 2
10/2 = sqrt(x-5) =>
5 = sqrt(x-5) => square both sides
5*5 = x-5 => add 5 to both sides
5*5 + 5 = x
25 + 5 = x
x = 30
Thank you,
MrB
Pakal triangle is hard to do with long things, but it is easy up to like 5 degree
we look at the row for 5th degree (6th row)
the sequence is
1,5,10,10,5,1
that is the coeficients
for (a+b)^5 that is
1a^5b^0+5a^4b^1+10a^3b^2+10a^2b^3+5a^1b^4+1a^0b^5
see the exponents each time add to 5
so
1x^5(-5)^0+5x^4(-5)^1+10x^3(-5)^2+10x^2(-5)^3+5x^1(-5)^4+1x^0(-5)^5=
x^5-25x^4+250x^3-1250
I'm guessing it represents the perimeter.
