A(bx − c) ≥ bc, implies (bx − c) ≥ bc /a and then bx ≥ bc/a + c, x<span>≥ c/a +c/b
so the solution is </span><span>3. [c/a + c/b, infinity)</span>
Ho ho ho, lets get this party started
ok so I'm just really excited to use this stuff that I just learned
so
multiplicites
if a root or zero has an even multilicity, the graph bounces on that root
if the root or zero has an odd multiplicty, the graph goes through that root
so
roots are
-1
2
4
multiplicty is how many times it repeats
2 has even multiplity
we just do 2 is odd and 1 is even so
for roots, r1 and r2, the facotrs would be
(x-r1)(x-r2)
so
(x-(-1))^1(x-2)^2(x-4)
(x+1)(x-2)^2(x-4)
this is a 4th degre equaton
normally, it is goig from top right to top left
it is upside down
theefor it has negative leading coefient
y=-k(x+1)(x-4)(x-2)^2
In order to find the number of hats, multiply all the choices together, disregarding size:
4 * 4 * 7 <-- 4 styles * 4 stitches * 7 yarns
= 16 * 7 = 112
112 different hats can be made.
Just so u know, ur output value is f(x) and ur input value is x
f(x) = -2x^2 - 3x + 5....when ur input value(x) is -3
f(-3) = -2(-3^2) - 3(-3) + 5 =
f(-3) = -2(9) + 9 + 5
f(-3) = -18 + 14
f(-3) = -4 <==
Answer:
Step-by-step explanation:
Given
Yearly Cost for members; 
Required
Determine the yearly cost for non members
From the question, we understand that:
<em>A non member pays $0.20 for each game;</em>
Yearly cost is then calculated as thus;
y = Amount * Number of Games
Where:
Amount = $0.2 and Number of Games = x
Convert 0.2 to fraction
Divide the numerator and the denominator by 2
<em>Hence, the yearly cost for non members is </em>
