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3 years ago

11

Can somebody please help me

1 answer:

seraphim [82]3 years ago

6 0

Answer:

3) 36

2) 55

Step-by-step explanation:

We have to place the numbers in order from least to greatest.

3) 23, 26, 26, 32, 32, 36, 50, 52, 52, 52, 59,

4) 40, 42, 49, 54, 62, 67, 67, 70, 95

The more is the the difference between the lowest and highest values.

95-40=55

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Please help quick i get brainliest

IrinaK [193]

Question 1:
If you know that 3 squares equals 7 circles and the squares are being counted by 3, then continue to count the circle by 3 until you get 25 circles.
Example:
3 Squares : 7 circles
4 squares : 10 circles
5 squares : 13 circles
6 squares : 16 circles
7 squares : 19 circles
8 squares : 22 circles
9 squares : 25 circles Add 9 squares to 3 = 12
You will have 12 squares when continues to count all the way to 25 circles.

Question 2:
Just multiply 14 by 28 and you will get your answer. Which is 3. The missing digit's 3. You can then check by doing 392 ÷ 14 to see if you get 28.

6 0

3 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

<u>iii) Add the volume of the two frustrums</u>

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

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Answer: 6 and 6

Step-by-step explanation:6x6=36

6+6=12

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