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umka2103 [35]
3 years ago
10

Help with five plz!!!!!

Mathematics
1 answer:
frutty [35]3 years ago
5 0

Answer:

No there cannot be the same number of stickers on each page.

Step-by-step explanation:

If you want to find out how many stickers need to be in every page to be even you would add all the stickers up. 6+6+9+10+11= 42. Take the 42 and divide it by 5 to see how many stickers would go in each page. This will give you 8.4. However since this number is a decimal it can't be split evenly in whole stickers for each page. Meaning that it wouldn't be possible for each page to have a evenly distributed number of stickers per each page.

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How many elements are in the union of five sets if the sets contain 10,000 elements each, each pair of sets has 1000 common elem
masya89 [10]

Answer:

40951

Step-by-step explanation:

Using the principles of inclusion - Exclusion

Where C(n,r)=n!/(n-r)!r!

Total elements in the five sets including number repetition is given as (10000)×C(5, 1) =10000× 5!/(5-1)!1!=10000×5=50000

Total Number of elements in each pair including number repetition of sets is given as

=(1000) × C(5, 2) =10000

Number of elements in each triple of sets is given as

=(100) × C(5, 3) =1000

Number of elements in every four sets

=(10) × C(5, 4)=50

Number of elements in every one set

(1) × C(5, 5)=1

Therefore total number of unique elements=50000-10000+1000-50+1

=40951

5 0
3 years ago
The quantity a varies directly with b and c and inversely with dThe quantity d is tripledWhich of the following must be true for
EastWind [94]

Answer:

We have the next relation:

A = (b*d)/c

because we have direct variation with b and d, but inversely variation with c.

Now, if we have 3d instead of d, we have:

A' = (b*(3d))/c

now, we want A' = A. If b,c, and d are the same in both equations, we have that:

3bd/c = b*d/c

this will only be true if b or/and d are equal to 0.

If d remains unchanged, and we can play with the other two variables we have:

3b'd/c' = bd/c

3b'/c' = b/c

from this we can took that: if c' = c, then b' = b/3, and if b = b', then c' = 3c.

Of course, there are other infinitely large possible combinations that are also a solution for this problem where neither b' = b or c' = c

6 0
3 years ago
A sine function increasing through the origin can be used to model light waves. Violet light has a wavelength of 400 nanometers.
andrew11 [14]

Answer:

(2) (100, 300)

Step-by-step explanation:

One wave can be divided on 4 parts, as show on the picture.

Since a length of one whole wave equals 400, a length of each part is 400:4=100 nanometers.

The height of the wave is decreasing on the second and on the third part, hence over the interval (100, 300).

3 0
3 years ago
A bill for Rs 3235 was drawn on Feb. 3, 1998 at 6 months and discounted on March 3, 1998 at 8 % p.a. Find the value of the bill
Fantom [35]

Answer:

RS 3127.166

RS 107.834

Step-by-step explanation:

6 month bill :

Drawn date = Feb, 3

Discounted date = March 3,

Difference = 1 month

6 month - 1 month = 5 months = 5 * 30 = 150 days

Value of the bill :

P(1 - rate * time)

3235(1 - 0.08 * 150/360)

3235(1 - 0.08 * 0.4166666)

3235(1 - 0.0333333)

3235(0.9666666)

= Rs3127.1666

Value of bill = 3127.166

Bankers gain = 3235 - 3127.166 = 107.834

6 0
3 years ago
What are the values of the variables in the triangle below?
harkovskaia [24]

Answer:

D) x=9 ; y=3√(3)

Step-by-step explanation:

Sin(30)=\cfrac{y}{6\sqrt{3} }

\cfrac{18y}{6\sqrt{3}}=18\sin \left(30^{\circ \:}\right)

\sqrt{3}y=9

\cfrac{\sqrt{3}y}{\sqrt{3}}=\cfrac{9}{\sqrt{3}}

y=3\sqrt{3}

~

Cos(30)=\cfrac{x}{6\sqrt{3} }

\cfrac{18x}{6\sqrt{3}}=18\cos \left(30^{\circ \:}\right)

\sqrt{3}x=9\sqrt{3}

\cfrac{\sqrt{3}x}{\sqrt{3}}=\cfrac{9\sqrt{3}}{\sqrt{3}}

x=9

Therefore, x=9 and y=3√(3)

~

8 0
2 years ago
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