Which equation is solved for the variable *c*. a(b-c) =d A.) c= ab-d/b B.) c= ab+d/a C.) c= d-ab/a D.) c= ab-d/a Reassurance help? XD
a(b-c) =d ab - ac = d ad - d = ac now divide both sides by a
I'm not given any variables, I don't get it q.q
a(b-c) =d (divide a on both sides) b-c = d/a (subtract b on both sides) -c = (d/a) - b (reverse signs to get a positive c) c = -(d/a) +b a(b-c) =d (multiply a into parenthesis) ab - ac = d (subtract ab on both sides) -ac = d - ab (divide both sides by -a) c = -(d/a) + b
sorry typo last line should be ab - d = ac
Doesn't seem to match any of the choices though, sure you entered them right?
So confused.
And yes, my lest question had a typo on the actual test apparently. So this could be incorrect, I'm not sure
Basically you're just trying to get C by itself by using algebra
But I don't know how to ._. That's what I've been trying to figure out
Alright...so you're starting with a(b-c) =d
looks like there's one a too many in option D
a(b-c) =d To get the "a" to the other side, you divide by it on other sides and you're left with (b-c) = d/a
So then I just keep doing that?
(b-c) = d/a To get the "b" on the other side, just subtract it on both sides and you're left with -c = (d/a) -b
This is so frustratingly confusing to me, I'm sorry. Bad at math in general
-c = (d/a) -b Since you want "C" to be positive, you reserve the signs on both sides and you get c = -(d/a) + b
So, there's 2 sides to this equation a(b-c) =d a(b-c) on the left side and "d" on the right side and you want to get c by itself
Step 1 - you want to get ride of the "a" on the left side a(b-c) -to do that, you want to divide the left side by "a" -but by doing that, you also have to divide the right side by "a" (what you do to one side, you also have to do to the other side So...you're left with (b-c) on the left side and (d/a) on the right side
Oh, so then it's also like rewriting the equation each time till you get to the final "product"..?
yep
So then... (b-c) = (d/a) Just wanna make sure I'm following correctly
Step 2 - so now you have (b-c) on the left side (d/a) on the right side - Now you want to get ride of the "b" on the left side. - To do that, you have to subtract "b" from the left side, which you also have to do to the right side You can break the parenthesis on the left side b - c - b = -c on the left side and (d/a) - b on the right side
and yes, that's correct
How do you get negative c? Ugh I don't understand at all.
Step 3 (-c) on left side by itself, but it's negative currently ((d/a) - b) on the right side Since you want an equation to find the positive value of "c", you want to reverse the signs on both sides to give you c (positive) on the left side and (-(d/a) + b) on the right side put it all together c = -(d/a) + b
I don't have an option with anything close to that. D;
During step 2, the "C" is already negative (b-c) the C is negative, while the B is positive (note the subtraction sign in front of C)
Oh..? So when you're taking away the b, you aren't taking the sign with it...??
Yeah, I looked through the answers and none of them seem to match the answer I got. Either I completely messed up, or there's a mistype
I'm thinking my whole quiz is messed up. I was asked to solve for "variable a" in a previous question and there was no a anywhere.
Nop, the negative sign (-) stays with the C The B is positive You're basically adding a negative B (-b) to both sides
Oh ok, makes a bit more sense now
C.) c= d-ab/a seems closed to the answer if only: C.) c= (d-ab)/(-a)
I guess I'll go with that one lol. I'll tell you what it says afterwards if you'd like.
And thank you like so much for the help u-u I've been stumped on all these for days
actually...D could work also, but it needs a parenthesis
Oh, I'll just go with C.
D.) c= ab-d/a to c = (ab-d)/a idk...D seems to be closer since it's only missing a parenthesis, while C is missing parenthesis and a negative sign
Well...either way, if it is a mistype, it's on whoever typed/wrote that test
Okay, I'll have to ask my teacher about it afterwards. This seems to happen a lot, actually.
Alright...good luck
D was correct. I also asked a friend about it and he said that parenthesis should've been included.
@porkbuns
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