# Posts by Mithra

Total # Posts: 31

**algebra**

Let the # = x Four time the # = 4(x) The # increased by 6 4x + 6 = 94

**math**

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 time = distance / speed ...

**Math**

What “P” stands for?

**algebra**

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!

**math**

No! If you take the (-9) over on the left-hand side of the equation the sign will change to positive.

**algebra**

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)

**math**

x + 1/6 = 1/3 x = 1/3 - 1/6 x = ? Finish it from there!

**math**

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

**Algebra**

3x^2 - 27 3(x^2 - 9) 3[(x - 3)(? + ?)]

**math**

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

**algebra**

x^2 - 81 (x - 9)(x + 9)

**Math**

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, [(72 + y) / y] = 4 72...

**Math**

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.

**math algebra**

A rectangle, the opposite sides are parallel and equal of lenght. Length = (x + 12)cm Width = (x)cm

**algebra**

Sorry, I made a mistake. Substitute x = 80 in equation 1 or equation 2 and fine for y.

**Algebra**

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

**Algebra**

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

**Algebra**

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 -25y^2 9x^2 -25y^2 = (3x...

**Math**

(8x-3y)^2 (8x - 3y) (8x - 3y) Try finish it from there...

**Math**

(2x-4y)(2x+4y) 4x^2 + 8xy - 8xy - 16y^2 4x^2 - 16y2

**algebra**

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.

**Algebra**

-12 = w/-4 Multiply both sides by -4, we get: (-4 X -12)= -4w/-4 48 = w

**intermediate algebra**

(7x^2 – y) (7x^2 + y) 49x^4 + 7x^2 y -7x^2 y – y^2 49x^4 –y^2

**intermediate algebra**

(5x -6y) (5x+6y) 25x^2 + 30xy -30xy -36y^2 25x^2 - 30y^2

**intermediate algebra**

9y^2(y^4 + 8y^3 - y^2 + 8y + 7) 36y^6 72y^5 - 9y^4 + 72y^3 + 63y^2

**algebra1**

½ x – 6 = 5 ½ x = 6 + 5 = 11 X = 11 ÷ ½ X = 22

**math**

12 – 8 (3x – 5) > -7x + 29 12 – 24x + 40 > -7x + 29 24x -7x > 12 + 40 – 29 17x > 23 X > 23/17

**math**

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

**algebra**

(y^2 - 6y + 8) (y + 8) y^3 + 8y^2 -6y^2 -48y + 8y + 64 Y^3 + y^2 - 40y + 64

**algebra**

(x + 2)^2 (x + 2) (x + 2) X ^2 + 2x + 2x + 4 X^2 + 4x + 4

**math**

A n C {10, 20}

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