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# Alot of maths problems?

July 21, 2019 / By Dalinda
Question: i need to know alot about 40, is it; 1. odd or even 2.composite or prime 3.factors and multiples (up to at least 10) i know this is alot so if you like maths, then please do it :) thankyou so much omg u dont know how much i will love the person who answers this! <3 :) xoxoxoxoxoox 4.other numeral systems 5.square or triangular number (or neither) 6.fibonacci or pascals triangle or any other intresting facts i need these answers for these numbers; 40 11 2000 27 4 6 7 8 163 102 11 yeah bout that.... yes this is homework but give me a break will ya? and yea i know its even, but its just to show u the questions.

## Best Answers: Alot of maths problems?

Blessing | 4 days ago
Yeah, someone didn't do their homework, did they? Are you seriously asking if 40 is an odd number or even? What grade are you in?
👍 156 | 👎 4
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We found more questions related to the topic: Ask questions about math homework

1) 1600 = 16 * 100 = 2^4 * 10 * 10 = 2^6 * 5^2 2) 30m:40m 3) (4x + 16)/4 = 4(x + 4)/4 = x + 4 4) x/5 - 1 < 4 x/5 < 5 x < 25 5) For this one, you basically want the amount of internal cubes to be larger than the amount of external cubes. Internal cubes = (x - 2)³ External cubes = x³ - (x - 2)³ You want (x - 2)³ > x³ - (x - 2)³ If we work out the point at which point they equal, then the answer we want will be the integer above. (x - 2)³ = x³ - (x - 2)³ x³ - 6x² + 12x - 8 = x³ - (x³ - 6x² + 12x - 8) x³ - 6x² + 12x - 8 = 6x² - 12x + 8 x³ - 12x² + 24x - 16 = 0 If you solve this, you will get a number of x = 9.69.... So the smallest block in which the internal cubes outnumber the external cubes is 10*10*10
Hi, 1) Write 1600 as a product of prime numbers in index form. 1600 = 2^6*5^2 2) Divide 70m into the ratio 3:4 ¾ ÷ 70m = ¾ × 1/(70m) = 3/(280m) 3) Factorize and Simplify: 4x + 16 / 4 4x + 16 ---------- = ...4 4(x+4) --------- = x + 4 <== answer ..4 4) Solve the inequality x/5 - 1<4 - Is the answer x<15? No x/5 - 1 < 4 Add 1. x/5 < 5 Multiply by 5. 5(x/5) < 5(5) x < 25 <== answer 5) This is a bit of a tricky one.... - If twenty-seven cubes are glued together to form a 3*3*3 block, one cube will be completely hidden. In a 4*4*4 block, eight of the sixty-four cubes are completely hidden. What are the dimensions of the smallest block in which more than half the cubes are hidden? 10*10*10 is the smallest block in which more than half the cubes are hidden. In it you can see 4 sides for walls of 10x 9 blocks( only 9 so you don't repeat the corner column on both sides) plus 8 x 8 on the top center and 8 x 8 on the bottom center for a total of 488 on the outside. 8*8*8 or 512 are on the inside, which is more than half. A quick way to do this on a graphing calculator is to enter y = (x - 2)^3/x^3 and look at the table to see the first value greater than .5 for y. I hope that helps!! :-)