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How can you use a 2s fact and 1s fact to find 3x8?
(1 x 3 ) x ( 2 X 4) = 24
How can you use a 2s fact and a 1s fact to find 3x8?
3x8 with1s and 2s fact looks like this: =(2x8)+(1x8) =16+8 =24
Ralph is 3 times as old as Sara In 6 years Ralph will be only twice as old as Sara will be then Find Ralph's age now?
1) 3S = R 2) R+6= 2 (S+6)= 2S+12 ----> R=2S+6 1,2 )--> 3S = 2S+6 ---> S=6 & R= 3 x 6 = 18 Ralph is 18 & Sara is 6 .
How do you find the factor tree for 75?
75 25, 3 5, 5, 3
Which double fact can help you find the sum 4 3?
6+7
How can use a 2s fact and a 1s fact to find 3 x 8?
To find 3 x 8 using a 2s fact and a 1s fact, you can break down 8 into manageable parts. For instance, you know that 3 x 2 = 6 (a 2s fact) and 3 x 1 = 3 (a 1s fact). Then, you can combine these to find 3 x 8 by calculating 3 x (2 + 2 + 2) = 3 x 2 + 3 x 2 + 3 x 2, which equals 6 + 6 + 6 = 24. Alternatively, you can also use 3 x 4 (2s fact) and double it to get 24.
How can you use 2s fact and 1s fact to find 3 times 8?
3 times 8 it's only 24 always
Factor the poynomial twice first use a common factor with a positive coefficient and then use a common factor with a negative coefficient -2s4 4s3-10s?
2s(-s^3 + 2s^2 - 5) -2s(s^3 - 2s^2 + 5)
How do you factor 4s2-9?
4s2 - 9 can be expressed by using the identity: a2 - b2 = (a-b)(a+b) Therefore, 4s2 - 9 = (2s)2 - 32 = (2s-3)(2s+3)
How can you use a 2s fact and 1s fact to find 3 times 8?
3 * 8 = (2 + 1)*8 = (2 * 8) + (1 * 8)
How can you use a 2s fact and 1s fact to find 3x8?
(1 x 3 ) x ( 2 X 4) = 24
What is the factor of 2s2-s-15?
2s2-s-15 = (2s+5)(s-3)
What is 9r squared-4s square over 9r plus 6s?
9r2-4s2/9r+6sIt looks like you can factors the numerator(3r + 2s)(3r - 2s) [This is the factored form of 9r2-4s2]Put this back into the equation(3r+2s)(3r-2s)/9r+6sYou can also factor the denominator3(3r+2s)Put this back into the equation(3r+2s)(3r-2s)/3(3r+2s)You can cancel out the 3r+2s on top and bottom because they are the same they equal 1. Therefore your final answer is3r-2s over 3You could go further and say this is...r-(2/3)seither one is correct
How you find the least common denominator?
For example, factor each denominator into prime factors. Then multiply all the prime factors, eliminate duplicates BUT ONLY if they appear in the factorisation of each of the numbers. NOT if a factor is duplicated in the same number. eg 36 = 2*2*3*3 and 48 = 2*2*2*2*3 Then two 2s and one 3 are duplicated. Note that the only two of the 2s are considered duplicates even though there are 4 of them in the factorisation of 48. So the LCD is [2]*[2]*[3]*3*2*2*2*2*3*3 where the duplicates to be removed are shown in square brackets. = 144.
How do you determine the equation for the perpendicular bisector of the straight line joining the points s 2s and 3s 8s?
A = (s, 2s), B = (3s, 8s) The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s) Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3 Gradient of perpendicular to AB = -1/(slope AB) = -1/3 Now, line through C = (2s, 5s) with gradient -1/3 is y - 5s = -1/3*(x - 2s) = 1/3*(2s - x) or 3y - 15s = 2s - x or x + 3y = 17s
-4 plus 3 plus 2s 15?
If you mean: -4+3+2s = 15 then s = 8
How can you use a 2s fact and a 1s fact to find 3x8?
3x8 with1s and 2s fact looks like this: =(2x8)+(1x8) =16+8 =24
