One-Hundred-thirty-four
100+30+2
1$ d d d p p p p
CXXXII
C. p ‹ 12
x^(4) dx = x^(5) /5 [2,4] => 4^(5)/5 - 2^(5)/5 => 2^(10) / 5 - 2^(5) / 5 [2^(10) - 2^(5) ] / 5 = [1024 - 32] / 5 = 992/5 = 198.4
555 x 5 x 5 x 5 x 5 is five 5's multiplied by each other; when writing an exponent, use the form Bx, where 'B' is the base(the number being multiplied by itself), and 'x' is the exponent(how many times the base is being multiplied). Here, it would thus be written as 55.
13
5X + 5 = 25 subtract 5 from each side 5X + 5 - 5 = 25 - 5 5X = 20 divide both sides integers by 5 (5/5)X = 20/5 X = 4 -----------check in original equation by inserting 4 into X place 5(4) + 5 = 25 20 + 5 = 25 25 = 25 -----------checks
66 x 2 = 132
62*2
132 / 5 = 26.4
26.4
eleven dozens
The only name for 132 in English is one hundred (and) thirty-two.
Do your math.132/5=26.4.
132 over 5 as a mixed number = 262/5
132 is a composite number but 5 is not one of its factors
Five goes into 132 a total of 26 times. This can be calculated by dividing 132 by 5, which equals 26 with no remainder. In mathematical terms, this can be expressed as 132 ÷ 5 = 26.
5/12 and 5/11, same as 55/132 and 60/132 Now you decide
132 to 125.4 is a decrease of 5%.