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2 years ago

4) A bag of sweets has orange and red sweets in the ratio 2:3. The number of red sweets

Mathematics

2 answers:

mel-nik [20]2 years ago

8 0

Answer:

Solomon ate 4 sweets

Step-by-step explanation:

fraction of red sweets =3/5 of s,

where s is total number of sweets

51=3/5*s

51*5=3s

255=3s

s=255/3

s=85

85-51=34

there were 34 orange sweets

now there is

17/27=51/s

fraction of red sweets where s is total number of sweets

17s=51*27

17s=1377

s=1377/17

s=81

85-81=4

Solomon ate 4 orange sweets

Likurg_2 [28]2 years ago

6 0

Answer:

Step-by-step explanation:

17*3 = 51

10*3 = 30

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2 years ago

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A bookshelf is 42 inches long. How many books of length 1\frac{1}{2}1 2 1 inches will fit on the bookshelf? Explain your reaso

Blababa [14]

Answer: See explanation

Step-by-step explanation:

Your question isn't well written but I saw an identical question:

A bookshelf is 42 inches long. Each book is 1 1/4 inches wide. How many books will fit on the shelf?

For us to solve this, we would divide the length of the bookshelf by the width of each book. This would be:

= 42 ÷ 1 1/4

= 42 ÷ 5/4

= 42 × 4/5

= 33 3/5

Therefore, the number of books that will fit on the shelf is 33.

7 0

2 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

James has a 6-ft board that is 24 in. wide. He wants to cut it into five pieces to make a small bookshelf with two shelves. The

Elodia [21]

Answer:

width=x, length = y

(in ft)

3x + 2y = 6y>x

y = 1.5 , x= 1

1ft = 12 in 1.5ft = 18in

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2 years ago