Mithra
This page lists questions and answers that were posted by visitors named Mithra.
Questions
The following questions were asked by visitors named Mithra.
Answers
The following answers were posted by visitors named Mithra.
½ x – 6 = 5 ½ x = 6 + 5 = 11 X = 11 ÷ ½ X = 22
14 years ago
12 – 8 (3x – 5) > -7x + 29 12 – 24x + 40 > -7x + 29 24x -7x > 12 + 40 – 29 17x > 23 X > 23/17
14 years ago
X = 1/25 (22z – 9w) X = 22z/25 – 9w/25 X + 9w/25 = (22z/25) 25(x + 9w/25) = 22z 25x + 9w = 22z (25x + 9w) ÷ 22 = z
14 years ago
(x + 2)^2 (x + 2) (x + 2) X ^2 + 2x + 2x + 4 X^2 + 4x + 4
14 years ago
(y^2 - 6y + 8) (y + 8) y^3 + 8y^2 -6y^2 -48y + 8y + 64 Y^3 + y^2 - 40y + 64
14 years ago
A n C {10, 20}
14 years ago
(7x^2 – y) (7x^2 + y) 49x^4 + 7x^2 y -7x^2 y – y^2 49x^4 –y^2
14 years ago
(5x -6y) (5x+6y) 25x^2 + 30xy -30xy -36y^2 25x^2 - 30y^2
14 years ago
9y^2(y^4 + 8y^3 - y^2 + 8y + 7) 36y^6 72y^5 - 9y^4 + 72y^3 + 63y^2
14 years ago
-12 = w/-4 Multiply both sides by -4, we get: (-4 X -12)= -4w/-4 48 = w
14 years ago
Let siate age = x And Kim aged = x + 9 Since kim age is one more than 3 times siate age Therefore: x + 9 = 1 + 3x 9 - 1 = 3x - x 8 = 2x x = 4 (Siate age) Kim age is 1 more than 3 times Siate age: 1 + 3x = 13 Kim age.
14 years ago
(2x-4y)(2x+4y) 4x^2 + 8xy - 8xy - 16y^2 4x^2 - 16y2
14 years ago
(8x-3y)^2 (8x - 3y) (8x - 3y) Try finish it from there...
14 years ago
Here is an example: 9x^2 - 25y^2 ------------ 3x - 5y Factor the numerator, its a difference of perfect squares. 9x^2 - 25y^2 = (3x - 5y) (3x + 5y) (3x - 5y)(3x + 5y) (3x)(3x) = 9x^2 (3x)(5y) = 15xy (-5y)(3x)= -15xy (-5y)(5y)= -25y^2 9x^2 + 15xy - 15xy -2...
14 years ago
x + 6y = 12 --- Equation 1 -x + 7y = 1 --- Equation 2 13y = 13 y = 13/13 y = 1 sub y = 1 in equation 1 we get: x + 6y = 12 x + 6(1)= 12 x + 6 = 12 x = 12 - 6 x = 6 Hence, x = 6 when y = 1
14 years ago
8x - 3y = 94 ---- Equation 1 2x + 22 = y ---- Equation 2 -------------- Sub y = 2x + 22 in equation 1 8x - 3(2x + 22) = 94 8x - 6x - 66 = 94 2x = 94 + 66 2x = 160 x = 160/2 x = 80 Now sub x = 8 in eq 1 or eq 2 and find for y
14 years ago
Sorry, I made a mistake. Substitute x = 80 in equation 1 or equation 2 and fine for y.
14 years ago
A rectangle, the opposite sides are parallel and equal of lenght. Length = (x + 12)cm Width = (x)cm
14 years ago
Speed = distance / time 20 minutes = 20/60 = 1/3 hr Speed = 40 miles / (1/3) hr Speed = 120 mp/h Hence, his average speed returned home is 120 mile per hour.
14 years ago
Let the 1st # = x And the 2nd # = y Difference between the two # is = 72 x - y = 72 ---- Equation 1 Then, one of the # divided by the other # which = 4 x/y = 4 ----Equation 1 x - y = 72 -- eq 1 x / y = 4 -- eq 2 x = 72 + y -- eq 3 Sub eq 3 in eq 2 we get,...
14 years ago
x^2 - 81 (x - 9)(x + 9)
14 years ago
time = distance / speed time = 17 mile / 43 mph time = 0.39535 = (0.3935 x 60) = 23.72 minutes. Now, multiply 23.72 by 60 = ? seconds
14 years ago
3x^2 - 27 3(x^2 - 9) 3[(x - 3)(? + ?)]
14 years ago
x + 6y = 12 --- eq 1 -x + 7y = 1 --- eq 2 ------------ 13y = 13 y = 13/13 y = 1 Sub y = 1 in eq 1 or eq 2 and find for x
14 years ago
x + 1/6 = 1/3 x = 1/3 - 1/6 x = ? Finish it from there!
14 years ago
First race car driver: time = distance / speed time = 50 miles / 90 mph time = 0.55556 x 60 time = 33 minutes 20 seconds The first race car driver will took 33 minutes & 20 seconds to finish the race Fifteen minutes later a second race car drive start tim...
14 years ago
What “P” stands for?
14 years ago
4/5(2x - 3)> 12 (8/5)x - 12/5 > 12 (8/5)x > 12 + (12/5) (8/5)x > 72/5 x > (72/5)/(8/5) x > (72/5) x (5/8) x > ? Finish it from there!
14 years ago
No! If you take the (-9) over on the left-hand side of the equation the sign will change to positive.
14 years ago
Let the # = x Four time the # = 4(x) The # increased by 6 4x + 6 = 94
14 years ago
t =(-b}ã(b^2-4ac))/2a t =(-7}ã(7^2-4(15)(-2)))/(2(15)) t=(-7}ã(49 +120))/30 t=(-7}ã169)/30 t=(-7}13)/30 t=(-7+13)/30 = 6/30 = 1/5 t=(-7-13)/30 = -(20/30) = -(2/3) Therefore, t = 1/5 & -(2/3)
14 years ago
Ecell=Ecathode-Eanode =-0.44+0.74 =0.30V
8 years ago