Suppose you are fencing a rectangular dog pen. The pen is to be 15 feet long and 7 feet wide. How much fencing would you need?

Mathematics

2 answers:

WARRIOR [948]3 years ago

7 0

Answer:

Step-by-step explain to solve for the maximum area, the first derivative should be solved and equate it to zero, then solve for x.

A = 37x - x^2

dA / dx = 37 - 2x

equate dA / dx = 0

0 = 37 - 2x

2x = 37

x = 37 / 2

x = 18.5 ft

so the maximum area is

A = 37x - x^2

A = 37( 18.5) - 18.5^2

A = 342.25 sq ft

so the answer is letter B

Readme [11.4K]3 years ago

4 0

Answer:

44 feet

Step-by-step explanation:

15 + 15 + 7 + 7 = 44

We can take all 4 sides and add them together to find the perimeter (aka how much fencing you need)

hope this helps :)

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An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$