The equation of the parabola could be written as y-k = a(x-h)^2, where (h,k) is the vertex. Thus, y-(-3) = a(x+4)^2, or y+3 = a(x+4)^2.
The coordinates of one x-intercept are (-11,0). Thus, y+3 = a(x+4)^2 becomes
0+3 = a(-11+4)^2, so that 3 = a(-7)^2, or 3 = 49a. Therefore, a = 3/49, and the equation of the parabola becomes
y+3 = (3/49)(x+4)^2.
To find the other x-intercept, let y = 0 and solve the resulting equation for x:
0+3 = (3/49)(x+4)^2, or (49/3)*2 = (x+4)^2
Taking the sqrt of both sides, plus or minus 49/3 = x+4.
plus 49/3 = x+4 results in 37/3 = x, whereas
minus 49/3 = x+4 results in x = -61/3. Unfortunatelyi, this disagrees with what we are told: that one x-intercept is x= -11, or (-11,0).
Trying again, using the quadratic equation y=ax^2 + bx + c,
we substitute the coordinates of the points (-4,-3) and (-11,0) and solve for {a, b, c}:
-3 = a(-4)^2 + b(-4) + c, or -3 = 16a - 4b + c
0 = a(-11)^2 - 11b + c, or 0 = 121a - 11b + c
If the vertex is at (-4,-3), then, because x= -b/(2a) also represents the x-coordinate of the vertex, -4 = b / (2a), or -8a = b, or
0 = 8a + b
Now we have 3 equations in 3 unknowns:
0 = 8a + 1b
-3 = 16a - 4b + c
0 = 121a - 11b + c
This system of 3 linear equations can be solved in various ways. I've used matrices, finding that a, b and c are all zero. This is wrong.
So, let's try again. Recall that x = -b / (2a) is the axis of symmetry, which in this case is x = -4. If one zero is at -11, this point is 7 units to the left of x = -4. The other zero is 7 units to the right of x = -4, that is, at x = 3.
Now we have 3 points on the parabola: (-11,0), (-4,-3) and (3,0).
This is sufficient info for us to determine {a,b,c} in y=ax^2+bx+c.
One by one we take these 3 points and subst. their coordinates into
y=ax^2+bx+c, obtaining 3 linear equations:
0=a(-11)^2 + b(-11) + 1c => 0 = 121a - 11b + 1c
-3 = a(-4)^2 +b(-4) + 1c => -3 = 16a - 4b + 1c
0 = a(3)^2 +b(3) + c => 0 = 9a +3b + 1c
Solving this system using matrices, I obtained a= 3/49, b= 24/49 and c= -99/49.
Then the equation of this parabola, based upon y = ax^2 + bx + c, is
y = (1/49)(3x^2 + 24x - 99) (answer)
Check: If x = -11, does y = 0?
(1/49)(3(-11)^2 + 24(-11) - 99 = (1/49)(3(121) - 11(24) - 99
= (1/49)(363 - 264 - 99) = (1/49)(0) YES!
y = (1/49)(3x^2 + 24x - 99) (answer)
Answer:lol so true
Step-by-step explanation:
Answer:
y=t−1+ce
−t
where t=tanx.
Given, cos
2
x
dx
dy
+y=tanx
⇒
dx
dy
+ysec
2
x=tanxsec
2
x ....(1)
Here P=sec
2
x⇒∫PdP=∫sec
2
xdx=tanx
∴I.F.=e
tanx
Multiplying (1) by I.F. we get
e
tanx
dx
dy
+e
tanx
ysec
2
x=e
tanx
tanxsec
2
x
Integrating both sides, we get
ye
tanx
=∫e
tanx
.tanxsec
2
xdx
Put tanx=t⇒sec
2
xdx=dt
∴ye
t
=∫te
t
dt=e
t
(t−1)+c
⇒y=t−1+ce
−t
where t=tanx
Answer:
15u + 141 = 420
Step-by-step explanation:
So basically we are modelling how much the coach is spending. Each player is given a uniform and a basketball
We can use 'u' as the cost of a single uniform
We can use 'b' as the cost of a single basketball
Since there is 15 players we must give a ball and uniform to each so
15u + 15b = 420
The problem gives us the cost of each basketball: 9.40
so now 15u + 15(9.4) = 420
15u + 141 = 420
