He did. Are you asking to make sure?
Answer:
p = 25,20 cm
h = 34,32 cm
A(min) = 864,86 cm²
Step-by-step explanation:
Let call p and h dimensions of a poster, ( length , and height respectively) and x and y dimensions of the printed area of the poster then
p = x + 8 and h = y + 12
Printed area = A(p) = 384 cm² and A(p) = x*y ⇒ y = A(p)/x
y = 384 / x
Poster area = A(t) = ( x + 8 ) * ( y + 12 ) ⇒ A(t) = ( x + 8 ) * [( 384/x ) + 12 ]
A(t) = 384 + 12x + 3072/x + 96 A(t) = 480 + 3072/x + 12x
A(t) = [480x + 12x² + 3072 ] / x
A´(t) = [(480 + 24x )* x - (480x + 12x² + 3072]/x²
A´(t) = 0 [(480 + 24x )* x - 480x - 12x² - 3072] =0
480x + 24x² -480x -12x² - 3072 = 0
12x² = 3072 x² = 296
x = 17,20 cm and y = 384/17,20 y = 22,32 cm
Notice if you substitu the value of x = 17,20 in A(t) ; A(t) >0 so we have a minimun at that point
Then dimensions of the poster
p = 17,20 + 8 = 25,20 cm
h = 22.32 + 12 =34,32 cm
A(min) = 25,20 *34.32
A(min) = 864,86 cm²
It would be 8 inches for 1 day. Take 80/10=8
Answer:
After 7 hours will be 1.95489493x10^12 viruses
Step-by-step explanation:
If the virus spread with 19% per hour after one hour it will increase 57 and continue in the 300+((300*0.19)^7) form, we just need to caculate the form
Hey, Imherkk!
You can check by using a calculator.
Another way is doing 17*6.
If you get 102(you do) then you're answer is right.
I hope this helps;)
