Answer:
A. Cylinder + cone
<u>Volume is the sum of volumes:</u>
- V = Vcon + Vcyl = 1/3πr²h₁ + πr²h₂
- V = 1/3π*9²*12 + π*9²*120 = 31554.2 cm³
<u>Surface area of cone:</u>
- A = A=πr(r+√(h₁²+r²)) = π*9(9 + √(9²+12²)) = 678.6 cm²
<u>Surface area of cylinder minus bases:</u>
- A = 2πrh₂ = 2π*9*120 = 6785.8 cm²
<u>Total surface area:</u>
- 678.6 + 6785.8 = 7464.4 cm²
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B. Cube+ pyramid
<u>Volume:</u>
- V = a³ + (1/3)a²h = a³ + (1/3)a²√(l²-(a/2)²)
- V = 8³ + (1/3)8²√(10²-4²) = 707.5 cm³
<u>Surface area of pyramid:</u>
- A = a² + 2al = 8² + 2*8*10 = 224 cm²
<u>Surface area of cube minus bases:</u>
- A = 4a² = 4(8²) = 256 cm²
<u>Total surface area:</u>
19 comes next in the series, followed by two 21s.
Answer: 8X - 2 I believe
Explanation: If you add 3X + 5X you get 8X since you can’t add them with the whole numbers without a variable. After that you add 4 + -6 (4 - 6) either that or its 8X + 10
The correct answer here I believe would be
x > or equal to 1
The dot is on one so it would be x blank one
and then the arrow is going towards the right so it would be greater
The circle is closed so it would be or equal to
In the case that the circle or “dot” on the 1 is OPEN the answer is x > 1
THE IMAGE IS BLURRY SO IF IT IS ACTUALLY AN OPEN CIRCLE the answer would be x > 1
I hope this helps! Please mark as brainliest if my answer was the MOST helpful <3
Answer:
No
Step-by-step explanation:
Given: There are 4 prime numbers between 10 and 20 i.e, 11,13,17, and 19.
To verify : If are there always the same number of prime numbers between 2 consecutive multiples of 10
Solution:
No, it is not so that there always the same number of prime numbers between 2 consecutive multiples of 10.
For example:
50 and 60 are multiples of 10 .
Prime numbers between 50 and 60 are 53 and 59 .
i.e, there are two prime numbers between 50 and 60.
Therefore, this contradicts the statement that there always the same number of prime number between 2 consecutive multiples of 10.