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What is the prime factorization of 3136?
PF of 3,136 = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 or 26 x 72
What is the prime factorization of 1344?
1344 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 7 = 26 x 3 x 7.
What is the circumcenter of the triangle with the points (23) (5-3) and (92)?
The circumcenter is at (137/26, 23/26) = (5 7/26, 23/26) The circumcenter is where the side perpendicular bisectors meet; this can be calculated by finding the equations of the perpendicular bisectors and solving the simultaneous equations that result (only 2 needs be solved, the third can be used as a check): Let the three points of the triangle be A (2, 3), B (5, -3), C (9, 2) The sides are AB, BC, AC with perpendicular bisectors AB', BC', AC' respectively; then: AB has midpoint ((5+2)/2, (-3 + 3)) = (7/2, 0) AB has gradient (-3 - 3)/(5 - 2) = -6/3 = -2 → AB' has gradient 1/2 → AB' has equation y - 0 = 1/2(x - 7/2) → y = 1/2(x - 7/2) BC has midpoint ((9 + 5)/2, (2 + -3)/2) = (7, -1/2) BC has gradient (2 - -3)/(9 - 5) = 5/4 → BC' has gradient -4/5 → BC' has equation y - -1/2 = -4/5(x - 7) → y = 4/5(7 - x) - 1/2 AC has midpoint ((9 + 2)/2, (2 + 3)/2) = (11/2, 5/2) AC has gradient (2 - 3)/(9 - 2) = -1/7 → AC' has gradient 7 → AC' has equation y - 5/2 = 7(x - 11/2) → y = 7(x - 11/2) + 5/2 Solving for point of intersection of AB' and AC': 1/2(x - 7/2) = 7(x - 11/2) + 5/2 → x - 7/2 = 14x - 77 + 5 → 13x = 72 - 7/2 = 137/2 → x = 137/26 (= 5 7/26) Using AB' y = 1/2(137/26 - 7/2) → y = 1/2(137/26 - 91/26) → y = 23/26 Checking using BC': y = 4/5(7 - x) - 1/2 → y = 4/5(7 - 137/26) - 1/2 → y = 4/5(182/26-137/26) - 13/26 → y = 4/5(45/26) - 13/26 → y = 36/26 - 13/26 = 23/26 as expected.
What is the prime factors of 100 x 26 x 49 to the tenth power?
2, 5, 7 and 13
What is the prime factorization chart 26 through 50?
2 x 13 = 26 3 x 3 x 3 = 27 2 x 2 x 7 = 28 29 2 x 3 x 5 = 30 31 2 x 2 x 2 x 2 x 2 = 32 3 x 11 = 33 2 x 17 = 34 5 x 7 = 35 2 x 2 x 3 x 3 = 36 37 2 x 19 = 38 3 x 13 = 39 2 x 2 x 2 x 5 = 40 41 2 x 3 x 7 = 42 43 2 x 2 x 11 = 44 3 x 3 x 5 = 45 2 x 23 = 46 47 2 x 2 x 2 x 2 x 3 = 48 7 x 7 = 49 2 x 5 x 5 = 50
What is the prime factorization of 3136?
PF of 3,136 = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 or 26 x 72
What are the factors of 26 and 91?
2 x 13 = 26 7 x 13 = 91
What is the prime factorization of 1344?
1344 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 7 = 26 x 3 x 7.
What gets multiplied to make 182?
1 x 182, 2 x 91, 7 x 26, 13 x 14, 14 x 13, 26 x 7, 91 x 2, 182 x 1
What is the next number 2 7 26 101 400?
2 x 4 -1 =7 7 x 4 - 2 = 26 26 x 4 - 3 = 101 101 x 4 - 4 = 400 400 x 4 - 5 = 1600 - 5 = 1595 So, the next number is 1595
What is the LCD of 21 14 26?
The least common denominator (LCD) is 1.
The sum of twice a number and seven?
2x + 7 = 33 2x = 33 - 7 2x = 26 X = 26/2 X = 13
What is 364 divisible by?
1, 2, 4, 7, 13, 14, 26, 28, 52, 91, 182, 364.
What is 7 percent of 26?
Let's assume X% of 26 is 7. Hence, 26*X/100=7 --> X=7*100?26=26.9= 27 (approx.)
What is 4 and one half times 7 over 13?
9/2 x 7/13 = 63/26 or 2 and 11/26
Which is not a factor of 56 A 4 B 8 C 14 D 26?
56 = 2 x 2 x 2 x 7 → Its factors are: 1, 2, 4, 7, 8, 14, 28, 56 → (D)26 is not one of them.
What is the circumcenter of the triangle with the points (23) (5-3) and (92)?
The circumcenter is at (137/26, 23/26) = (5 7/26, 23/26) The circumcenter is where the side perpendicular bisectors meet; this can be calculated by finding the equations of the perpendicular bisectors and solving the simultaneous equations that result (only 2 needs be solved, the third can be used as a check): Let the three points of the triangle be A (2, 3), B (5, -3), C (9, 2) The sides are AB, BC, AC with perpendicular bisectors AB', BC', AC' respectively; then: AB has midpoint ((5+2)/2, (-3 + 3)) = (7/2, 0) AB has gradient (-3 - 3)/(5 - 2) = -6/3 = -2 → AB' has gradient 1/2 → AB' has equation y - 0 = 1/2(x - 7/2) → y = 1/2(x - 7/2) BC has midpoint ((9 + 5)/2, (2 + -3)/2) = (7, -1/2) BC has gradient (2 - -3)/(9 - 5) = 5/4 → BC' has gradient -4/5 → BC' has equation y - -1/2 = -4/5(x - 7) → y = 4/5(7 - x) - 1/2 AC has midpoint ((9 + 2)/2, (2 + 3)/2) = (11/2, 5/2) AC has gradient (2 - 3)/(9 - 2) = -1/7 → AC' has gradient 7 → AC' has equation y - 5/2 = 7(x - 11/2) → y = 7(x - 11/2) + 5/2 Solving for point of intersection of AB' and AC': 1/2(x - 7/2) = 7(x - 11/2) + 5/2 → x - 7/2 = 14x - 77 + 5 → 13x = 72 - 7/2 = 137/2 → x = 137/26 (= 5 7/26) Using AB' y = 1/2(137/26 - 7/2) → y = 1/2(137/26 - 91/26) → y = 23/26 Checking using BC': y = 4/5(7 - x) - 1/2 → y = 4/5(7 - 137/26) - 1/2 → y = 4/5(182/26-137/26) - 13/26 → y = 4/5(45/26) - 13/26 → y = 36/26 - 13/26 = 23/26 as expected.
