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3 years ago

In a circle of radius 10 cm, a sector has an area of 40 (pi) sq. cm. What is the degree measure of the arc of the sector?

Mathematics

2 answers:

Rashid [163]3 years ago

6 0

Answer:

144°

Step-by-step explanation:

First, find the area of the circle, with the formula A = $\pi$ r²

Plug in 10 as the radius, and solve

A = $\pi$ r²

A = $\pi$ (10²)

A = 100 $\pi$

Using this, create a proportion that relates the area of the sector to the degree measure of the arc.

Let x represent the degree measure of the arc of the sector:

$\frac{40\pi }{100\pi }$ = $\frac{x}{360}$

Cross multiply and solve for x:

100 $\pi$ x = 14400 $\pi$

x = 144

So, the degree measure of the sector arc is 144°

Semmy [17]3 years ago

3 0

Answer:

<h2><u>Solution</u><u> </u><u>:</u><u>-</u></h2>

We know that

Area = πr²

Area = 3.14 × (10)²

Area = 314/100 × 100

Area = 314 cm²

Now

40π/314 = x/360

40 × 3.14/314 = x/360

125.6/314 = x/360

0.4 = x/360

0.4 × 360 = x

144 = x

$\\$

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Premier Bank is planning to have a new building constructed. They would like the length of the building to be 40 feet longer tha

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Let L, W, F, H, P represent, respectively, the length, width, number of floors, height, and perimeter of the building. Here are some equations that model the given information.

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We can rewrite these into a system of equations involving L, W, F.

Let's look first at the relation between the perimeter and height. The two versions of perimeter are equal to each other, so we have (after substituting for H) ...

... 3H = 2(L+W)

... 3(12F) = 2((W +40) +W) . . . . substitute for H and L

... 36F = 4W +80 . . . . . . . . . . . simplify

... 9F = W +20 . . . . . . . . . . . . . divide by 4

Now, we can look at the cost equations.

... 1180000 = 35000F + 10W(W+40) . . . . . substitute for the various cost terms and for L

... 118000 = 3500F +W² +40W . . . . . . . . . divide by 10

The two simplified equations in W and F are ...

9F = W +20
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You can solve the first for F and substitute for F in the second. Then you have a quadratic equation in W that can be solved by the usual methods. (Likely, the quadratic formula would be the best choice.) Here, we have elected to let a graphing calculator show the two solutions. One solution has negative dimensions, so is clearly infeasible.

There is one (nearly feasible) solution:

... W ≈ 180.788

... F ≈ 22.31

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A solution that comes about as close as possible is ...