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2 years ago

Jarron has a piece of chalk tied to the end of a

2 answers:

8 0

Answer: A 24 cm piece of string is cut into two pieces , one piece is used to form a circle and the other piece is used to form a square.

How should this string be cut so that the sum of the areas is a minimum .

Let x = the circumference of the circle

then

(24-x) = the perimeter of the square

Find the area of the circle

find r

2*pi*r = x

r =

Find the area of the circle

A =

A = sq/cm, the area of the circle

Find the area of the square

A = sq/cm the area of the square

The total area

At = +

Graph this equation, find the min

Min occurs when x=10.6 cm

cut string 10.6 cm from one end

Step-by-step explanation: Hope I help out alot (-: :-)

yan [13]2 years ago

7 0

Answer:

F- 153.86 in²

Step-by-step explanation:

Area = 3.14 × 7²

153.86 in²

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Round 13.9 to the nearest whole number

Rudik [331]

14
You are very welcome

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3 years ago

Read 2 more answers

the half-life of a new experimental drug in your bloodstream is 15 hours how long will it take for the drug to Decay to 60% of t

Ierofanga [76]

4 hours? i may be wrong

i did - 60 * 15

seems about right, since 50% would be 7.5 hours?

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2 years ago

PLEASE HELP!! Find the surface area of each figure to the nearest tenth. Show your work.

hram777 [196]

Surface area = 2(ab+bc+ac)
a=5.2 ft, b=2.4 ft, c=3.5ft

Surface area = 2(5.2 * 2.4 + 2.4 * 3.5 + 3.5 * 5.2) =
2(12.48 + 8.4 + 18.2) =
2 * 39.08 =
78.16 ≈ 78.2 ft² ← to the nearest tenth

7 0

3 years ago

Could these side lengths make a right triangle, 12m, 35m and 37m

natta225 [31]

Answer:

Yes

Step-by-step explanation:

Using Pythagorean theorem WHICH ONLY WORKS FOR RIGHT TRIANGLES,

a^2+b^2=c^2 where a and b are the two shortest legs.

12^2+35^2=c^2

144+1225=c^2

c^2=1369

c= $\sqrt{1369}$

c=37

3 0

2 years ago

Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

2 years ago