How to solve these math equations?
Remember your order of operations: 5x^2=405 Divide each side by 5 x^2 = 81 Take square root of each side x=9,-9 5x^2+8=58 Subtract 8 from each side 5x^2 = 50 Divide each side by 5 x^2 = 10 Take square root of each side x=SQRT(10) 5(x+1)^2=60 Divide each side by 5 (x+1)^2=60 Take square root of each side x+1 = SQRT(60) Subtract 1 from each side x = SQRT(6) - 1
How do you solve these math equations? 2/3+1/4t=6?
2/3+1/4t=6 1) The key is to isolate the variable by doing the opposite operation so that you can find out what the variable equals. First what you would do is subtract 2/3 from both sides. 1/4t = 5 1/3 2) Next divide each side by 1/4. This leaves t = 21 1/3 See if you can follow what I am doing for these next two problems: 1/3x + 2/5 = 4/15 + 3/5x - 2/3 1/3x +2/5 = -2/5 + 3/5x 1/3x = -4/5 + 3/5x -4/15x = -12/15 x = 180/60 x = 3 1/6 (3/4x-2) = -1/5 3/24x -2/6 = -1/5 1/8x -1/3 = -1/5 1/8x = 2/15 x = 16/15 x = 1 1/15
How do I solve this maths equation?
For expanding two brackets you need to use FOIL First Inner outer Last. For the left hand side(a^2+1)(c^2+1)First term in each bracket: a^2 * c^2 = a^2 c^2Outer two terms: a^2 * 1 = a^2Inner two terms: 1 * c^2 = c^2Last terms in each bracket: 1 * 1 = 1Add the terms together: a^2 c^2 + a^2 + c^2 + 1.For the right hand side note that (ac-1)^2 is the same as (ac-1) * (ac-1) and (a+c)^2 is the same as (a+c) * (a+c). Use FOIL on both these pair of brackets. The other answers show the rest.
Solve the equation - Math question.?
x3-343=0 +343=+343 ------------------- x3= 343 x=7
Could some one help me solve these math equations?
1/7 m=9 m=9*7/1 m=63 45=1/4 d 45*4/1=d 180=d n/3 = 16 n= 16*3 n=48 when you divide by a fraction, it is the same as multiplying by the reciprocal.
How to solve these math question?
t + b = 25 4b = t 4b + b = 25 5b = 25 b=5 5 bean plants and 20 tomato plants
How do I solve this math equation below?
B^2 = 6 - 2*5^1/2 + 6 + 2.5^1/2 +2(6-2*5^1/2)(6+2*5^1/2)^1/2 = 12 + 2(36-20)^1/2 = 12 + 2*4 = 20B = 20^1/2 = 2*5^1/2
How do you solve the mathematical equation y=mx+b?
There’s four variables there! It’s already solved for y. You could solve for x:[math]x=\frac{y-b}{m}[/math]You could solve for m:[math]m=\frac{y-b}{x}[/math]You could solve for b:[math]b=y-mx[/math]My guess is that you don’t actually want to do any of those things, and that you have a more general question about linear equations. Unfortunately, the “smart math people” need to know more about where your lack of knowledge lies. You’ll get a lot better a lot faster if you have someone go over this sort of thing with you, with a paper between you, practicing and discussing your weaknesses.You can definitely get such a tutor who will do their best to help you succeed. I recommend this company. At least give it a shot: Lynnwood WA SAT Tutoring(Yes, I work for this company. No, I cannot confirm the quality of this particular location. It just happens to be the location closest to you. I suspect it is quite a good one, since the lead teacher has a master’s in teaching, a fairly uncommon trait among C2 tutors.)