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3 years ago

What percent of 99.2 is 91

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

7 0

To find percent, you take 91 and divide it by 99.2 then multiply by 100.

(91/99.2) × 100 = 91.92%

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Find the area of the larger sector.

Wewaii [24]

Step-by-step explanation:

the formula for the area of a sector is

(x°(r^2)π)/360

with x being the angle

r bring the radius of the circle

(255(13.4)^2π)/360

399.6

Hope that helps :)

3 0

3 years ago

A company has determined that the number of items it sells varies inversely as the price of the item. If 20,000 items can be sol

mestny [16]

Let N be the number of items sold and p the price.
Since the variation is inverse, then the relation between N and p is:
$N=k\dfrac{1}{p}$
For N=20000 and p = $9.5, we get the formula:
$20000=k\dfrac{1}{9.5}\\p=20000\times9.5=190000$
If p = 8.75, then the number of items sold can be computed using the formula:
$N= 190000\dfrac{1}{8.75}=21714 \text{ total items}$

6 0

3 years ago

Find the degree of the monomial. 3a4b3

steposvetlana [31]

Answer:

degree of the monomial is the sum of the exponents of all included variables

Step-by-step explanation:

please mark me as brainlist please

3 0

2 years ago

I mark brainlest for anybody who answer's tis question ;-;

Zepler [3.9K]

Answer:

b~ 32

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago