The perimeter of a square equals 112 inches. how many inches would each side have to equal?

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

4 years ago

If the Lafita family deposits 8500$ ina savings account at 6.75% interest, compounded continuously, how much will be the account

julsineya [31]

A=pe^rt
A=8,500×e^(0.0675×25)
A=45,950.57

8 0

4 years ago

(8.166×10^8)×(1.515×10^9)

loris [4]

The answer is ....1.237149 x 10^18

4 0

3 years ago

Please help with these math questions part 2

NikAS [45]

5).
and
6).
The volume of a sphere is

   (4/3) (pi) (radius)³ .

In #5, the 'pi' is already there next to the answer window.
You just have to come up with the (4/3)(radius³).

Remember that the radius = 1/2 of the diameter.

7). The volume of a cylinder is

   (pi) (radius²) (height) .

Divide the juice in the container by the volume of one can,
to get the number of cans he can fill.

8). The volume of a cone is

   (1/3) (pi) (radius of the round bottom)² (height) .

He starts with a small cone, he then adds clay to it to make it higher.
The question is: How much clay does he ADD to the short one,
to make the bigger one ?

Use the formula to find the volume of the short one.
Use the formula again to find the volume of the bigger one.
Then SUBTRACT the smaller volume from the bigger volume.
THAT's how much clay he has to ADD.

Notice that the new built-up cone has the same radius
but more height than the first cone.
_______________________________________

Don't worry if you don't understand this.

The answer will be this number:

(1/3) (pi) (radius²) (height of the big one minus height of the small one).

6 0

3 years ago

Give the domain and range.

MA_775_DIABLO [31]

Answer:

Domain: {-3, 0 3}

Range: {0, 3}

Step-by-step explanation:

Recall that the Domain of a relationship associated with coordinate pairs of the form (x,y), is the set of all "x-values" given in the relationship.

The Range is the set of all "y-values" given in the relationship.

In our case, we have three coordinate points defining the relationship. These points have the following (x,y) coordinates (see attached image to see which coordinates are associated with each point):

(-3, 3), (0, 0), and (3, 3)

Therefore, the x values associated with these points are: -3, 0 and 3

therefore the Domain of the relationship is the set: { -3, 0, 3}

On the other hand, he y values associated with these points are: 3, 0, and again 3. So in fact there are just to distinct y-values: 0 and 3.

therefore the Range of the relationship is he set: {0, 3}

which corresponds to answer B.

6 0

3 years ago