Answer:
the third table
Step-by-step explanation:
Hope this helps and plz mark as brainliest!
Answer and explanation:
<em>To solve this, you must find what x equals</em>
x + 2x + 7 = 3x - 7 <em>Add together any numbers that can be added on their side</em>
3x + 7 = 3x - 7 <em>Then subtract 7 from both sides to get 3x alone</em>
3x = 3x - 14 <em> Then subtract 3x from both sides to get x</em>
<h3><em>*The x's cancel each other out so you must have typed the equation wrong. Please check to see any mistakes you may have made*</em></h3>
I'm not 100% sure, but I'm pretty sure it's 1/12, 1/12, 1/12, 1/12
2b 2a
----------------- + -----------------
(b+a)^2 (b^2 - a^2)
2b 2a
= ----------------- + -------------------
(b+a)(b+a) (b+a)(b-a)
2b(b - a) + 2a(b + a)
= ------------------------------------
(b+a)(b+a)(b-a)
2b^2 - 2ab + 2ab + 2a^2
= ---------------------------------------
(b+a)(b+a)(b-a)
2b^2 + 2a^2
= ------------------------
(b+a)(b+a)(b-a)
2(b^2 + a^2)
= ------------------------
(b+a)^2 (b-a)
Answer:
Numerator: 2(b^2 + a^2)
Denominator: (b+a)^2 (b-a)
Answer:
We have the next relation:
A = (b*d)/c
because we have direct variation with b and d, but inversely variation with c.
Now, if we have 3d instead of d, we have:
A' = (b*(3d))/c
now, we want A' = A. If b,c, and d are the same in both equations, we have that:
3bd/c = b*d/c
this will only be true if b or/and d are equal to 0.
If d remains unchanged, and we can play with the other two variables we have:
3b'd/c' = bd/c
3b'/c' = b/c
from this we can took that: if c' = c, then b' = b/3, and if b = b', then c' = 3c.
Of course, there are other infinitely large possible combinations that are also a solution for this problem where neither b' = b or c' = c
