how do i solve -3r+3r=-6(7+7r)-7(4-7r)
To answer a quadratic like this, st everything equal to zero. 5r^2+12r=-4r^2-4 9r^2+12r+4=0 Ask yourself what two numbers add to get 12 that multiply to get 9*4=36 1*36 no 2*18 no 3*12 no 4*9 no 6*6 yes (3r+2)(3r+2) The first and last multiply to get 6 and the outside and inside multiply to get 6. That's (3r+)^2
6p + 12q + 18r = 6 (p + 2q + 3r)
3.5
6, remainder 21
how do i solve -3r+3r=-6(7+7r)-7(4-7r)
9r = 3r + 6 subtract 3r from both sides 6r = 6 divide both sides by 6 r=1 ■
9r = 3r+6 Subtract 3r from both sides: 6r = 6 Divide both sides by 6 to find the value of r: r = 1
r+3+2r=21 to solver this equation you have to first simplify then solve. you have to combine same variables. 3r+3=21 -3 -3 you subtract three to get r by itself (3r=18)/3 you divide by three to get r by itself Answer r=6
No, it's simply 3r, because multiplication (r multiplied by 3) comes first in the order of operations.
-7p + 5r - 6p + 3r = -7p - 6p + 5r + 3r = -(7 + 6)p + (5 + 3)r = -13p + 8r
It does nothing. It has no capacity for doing anything. 14r + 6 * 3r - 2 = 14r + 18r - 2 = 32r - 2
The width and height of a 3R picture is traditionally 4"Ã?6". However, you can also resize this type of picture to fit your needs.
r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)
P(R on the third draw) = 3/5 = 0.60 = 60%To calculate the probability of this event you have to consider all the possible outcomes that give a Red ball on the third draw. They are four:[1R,2R,3R], [1R,2W,3R], [1W,2R,3R], [1W,2W,3R].The probability for these four events are:1.- P(1R,2R,3R) = (6/10)(5/9)(4/8) = 1/62.- P(1R,2W,3R) = (6/10)(4/9)(5/8) = 1/63.- P(1W,2R,3R) = (4/10)(6/9)(5/8) = 1/64.- P(1W,2W,3R) = (4/10)(3/9)(6/8) = 1/10Then the probability looked for is the sum of all four:P(R on the third draw) = 1/6 + 1/6 + 1/6 + 1/10 = 3/5 = 0.60 = 60%
The factors of 3 are 1 and 3 The factors of 12 are 1, 2, 3, 4, 6, and 12 The factors of 18 are 1, 2, 3, 6, 9, and 18
To answer a quadratic like this, st everything equal to zero. 5r^2+12r=-4r^2-4 9r^2+12r+4=0 Ask yourself what two numbers add to get 12 that multiply to get 9*4=36 1*36 no 2*18 no 3*12 no 4*9 no 6*6 yes (3r+2)(3r+2) The first and last multiply to get 6 and the outside and inside multiply to get 6. That's (3r+)^2