Answer:
$13.50
Step-by-step explanation:
if the total is $31.50, it is both the blouse and the skirt together, so uf you take away the blouse from the total ($31.50 - $18) then yiu are keft with the price of the skirt
Let
x ----------> the height of the whole poster
<span>y ----------> the </span>width<span> of the whole poster
</span>
We need
to minimize the area A=x*y
we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2
so
A(x)=x*y--------->A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4
find the derivative------> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]
for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in
(y)=[722/(x-4)]+2
y=[2x+714]/[x-4]-----> y=[2*13.87+714]/[13.87-4]-----> y=75.15 in
the answer is
<span>the dimensions of the poster will be
</span>the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in
20 and 10, 20 is 2 times 10 and if you add them together its 30
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
1 1/16 is the answer!! Hope this helps! <3 x
