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Orlov [11]
3 years ago
9

Find the mitsake: 1 2 3 4 5 6 7 8 9 10

Mathematics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

the mistake with the word mitsake

Step-by-step explanation:

You might be interested in
Share 3200 ml of water in the ratio 4:1:3​
Amiraneli [1.4K]

Answer:

1600 ml, 400 ml, 1200 ml

Step-by-step explanation:

Given ratios are: 4:1:3

Let the shares of the given ratios be 4x, x, 3x

\therefore \: 4x + x + 3x = 3200 \\  \\  \therefore \: 8x = 3200 \\  \\  \therefore \: x= \frac{ 3200}{8} \\  \\ \therefore \: x=400 \\  \\  \implies \\ 4x = 4 \times 400 = 1600 \\  \\ 3x = 3 \times 400 = 1200 \\  \\

8 0
3 years ago
Which decimal is the equivalent of 6/11 Repeated
liraira [26]
6/11 = 0.54_54 . . . . . the digits 54 are repeated indefinitely You can arrive at the digits 54 by multiplying the numerator (6) by 9. This will be the case for any proper fraction with 11 as a denominator.
7 0
4 years ago
What is the volume of a regular hexagonal prism with a side length of 4 and a height of 9? Please give the answer as square root
likoan [24]
The height of an equilateral triangle with side s is s*sqrt(3)/2
The area of an equilateral triangle is therefore s^2*sqrt(3)/4
The area of a hexagon with side s is 6 times the area of an equilateral triangle
=6*s^2*sqrt(3)/4
=3s^2*sqrt(3)/2
 
Here s=4, 
Area of base (hexagon) = 3*4^2*sqrt(3)/2=24sqrt(3)
Volume of prism
=(base area)*height
=24sqrt(3)*9
=216sqrt(3)

6 0
4 years ago
The ratio of girls to boys in a swimming club was 2:4 .There were 14 girls .How many total members were there in the club
dusya [7]

Answer:

42

Step-by-step explanation:

5 0
3 years ago
Please help me! in need of help due tomorrow!​
julsineya [31]

Answer:

Step-by-step explanation:

(1). A = π r²

A = 4 π

Area of shaded region is \frac{4\pi *260}{360} = \frac{26}{9} \pi or \frac{26\pi }{9}

(2). C = 4 π

In this case, the length of the bigger arc and the area of shaded region happen to be the same.

The length of the arc ADB is  \frac{26}{9} \pi or \frac{26\pi }{9}

3 0
2 years ago
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