Answer:
D
Step-by-step explanation:
observe, grouping by gcd not 1
28r² + 35ry – 4xr – 5xy
= 28r² – 4xr + 35ry – 5xy
= 4r(7r – x) + 5y(7r –x)
= (4r + 5y)(7r – x)
Area = length *width
A = x²
3x² = (x+10)(x+12)
3x² = x²+10x+12x+120
3x² = x²+22x+120
3x²-3x² = x²+22x+120-3x²
0 = -2x²+22x+120
Factor to solve for x:
-2(x²-11x-60)
(x+4)(x-15)
x = -4 & x = 15 but we can't have a negative length so we eliminate 4. The length of the square is 15 cm.
Let set C = {1, 2, 3, 4, 5, 6, 7, 8} and set D = {2, 4, 6, 8}.
g100num [7]
<span>1. If it is intersection then it SHOULD be included in both the sets right?
Now we know that odd numbers from 1-100 but the second set are multiples of 5 from 50-150! So we mainly need to look for common numbers which are ODD and are a MULTIPLE OF 5 BETWEEN 50 - 100!!
So
A={51,53,57,59,61......99}
B={55,60,65,70.......95} [We stop till 100 because set A has no such element]
So what is A ∩ B here?
A ∩ B = {All odd numbers and multiples of 5 between 50 - 100}
</span>
Answer:
21 Kg
Step-by-step explanation:
Divide 56 Kg by 8 and multiply the answer by 3 , that is
56 Kg ÷ 8 = 7 Kg , then
3 × 7Kg = 21 Kg
