Let's start by writing a system of linear equations:
c -> cookies
cb -> candy bars
(You can use any abbreviations to your preference)
Abby:
4 cookies
3 candy bars
$10.25 per bag
The equation would be:
4c+ 3cb = $10.25
Marissa:
2 cookies
7 candy bars
$14.75 per bag
The equation would be:
2c + 7cb = $14.75
So our linear equation system would be:
<span>4c+ 3cb = $10.25
</span><span>2c + 7cb = $14.75
I would try to get rid of one variable so I can solve for the other variable. In this case, it is easier to get rid of c since I can multiply the second equations by 2. Then it would subtract the two equations.
(2c + 7cb = $14.75) 2 = 4c + 14 cb = $29.50
4c + 3cb = $10.25
- 4c+14 cb = $29.50 (4c would get canceled.)
---------------------------------
-11 cb = - $19.25 (Divide by -11 to solve for cb)
</span> --------- -------------
-11 -11
cb = $1.75
Now we know cb (candy bar) cost, we would substitute this value into cb into one of the equations. It doesn't matter which equation you put it in. I will substitute it in the first equations.
4c + 3 (1.75) = $10.25
4c + 5.25 = $10.25 (Multiply 3 by 1.75)
-5.25 -5.25 (Subtract 5.25 on both sides)
4c = 5 (Divide by 4 on both sides to get c)
---- ---
4 4
c= 1.25
Check the work:
4(1.25) + 3(1.75)
= $10.25
2(1.25) + 7(1.75)
= $14.75
Total cost:
cookies = $1.25
candy bars = $ 1.75
Hope this helps! :)
Answer:
3x^2 -7x +2
Step-by-step explanation:
(15x^2 -35x +10)÷5
15/5 x^2 -35/5 x +10/5
3x^2 -7x +2
The additive inverse of a complex z is a complex number
so that
Finding
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Tags: <em>complex number additive inverse opposite algebra</em>
The answer for the question is 17/18
Substitute
, so that
. The integral is then equivalent to
Then transforming back to
gives
