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Sonbull [250]
3 years ago
13

There are 15 boxes with the same number of books in each box is a total number of books is 930 how many books are in each box

Mathematics
2 answers:
Ulleksa [173]3 years ago
8 0
930 divided by 15 = 62 books in each box
katovenus [111]3 years ago
4 0

Answer: 62 books

Step-by-step explanation: Divide 930 by 15. This gets you 62.

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jordan is twice as old as kevin. kevin is 4 times old as george. if george is 2 years old, then how old is jordan and kevin
umka2103 [35]
I think Jordan would be 16 and Kevin would be 8. Sorry if this is wrong...
3 0
3 years ago
Read 2 more answers
Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
Please answer this correctly as soon as possible I want genius or expert people to answer this correctly
Ede4ka [16]
6/54 times because there are 9 marbles and the orange marble is 1 out of those 9 marbles. If you multiply 6*9= 54 then you notice there is a 6/54 chance of picking any 1 marble out of the 54 times that you decide to pick marbles.
4 0
3 years ago
Step 1: 12x – 15 – 12x = 7x + 20
stealth61 [152]

Step-by-step explanation:

step 1. 12x - 15 - 12x = 7x + 20

step 2. 12x - 12x - 15 = 7x + 20 (grouping of terms)

step 3. -15 = 7x + 20 (adding like terms)

step 4. -35 = 7x (subtract 20 from both sides)

step 5. this step is incorrect.

step 6. -5 = x ( divide both sides by 7)

step 7. x = -5 (put the variable first).

5 0
3 years ago
Anyone, I need help... Just answer the 6 (c)....and also proper working.☺️
Ainat [17]

Answer:

(i) The area of the rabbit cage when the width is 5.2 m is 81.5 m²

(ii) The area of the rabbit cage if Wilson has 40 meters of wire mesh is 75 m²

Step-by-step explanation:

(i) The given relation of the area, A to the width P of the rabbit cage is A = 3·p²

The graph of the function between the values of 0 and 6 inclusive is found as follows;

A,              3·p²

0,               0

1,                1

2,               12

3,               27

4,               48

5,               75

6,               108

Please find attached the graph of A to 3·p²

From the graph, we have when the the width, p, of the rabbit cage = 5.2, the area, A ≈ 81.5 m²

The area of the rabbit cage when the width is 5.2 m = 81.5 m²

(ii) Also from the graph given that the total wire mess with Wilson = 40 meters, we have;

The formula for the perimeter of the cage = The formula for the perimeter of a rectangle = 2×length + 2×width

The formula for the perimeter of the cage = 2×3×p + 2× p = 8·p

Where the total length of the wire mesh available = 40 meters for the cage

The 40 meters of wire mesh will be used round the perimeter of the cage

∴ 40 m. = 8·p

p = 40/8 = 5 m.

At p = 5 m. the area is given as A = 75 m².

Therefore, the area of the rabbit cage if Wilson has 40 meters of wire mesh = 75 m².

5 0
3 years ago
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