PLS HELP ME WILL GIVE BRAINLIEST
1 answer:
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Choice C is the correct answer because
So in short,
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The mistake Jerry likely made was that he only cubed the 2 and didn't realize the 6 was part of that cubing process. It seems he didn't add first and decided to cube before adding.
This is probably what steps Jerry did
But as mentioned, those steps are incorrect because the 6 is part of the cubing operation. In other words, Jerry should have added the 6+2 first before cubing afterward (due to PEMDAS determining the order of operations).
Or you could think of it like this
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=
x=
x=
x=
x=
there are no real numbers that satisfy this
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=
x=
x=
x=
x=
there are no real numbers that satisfy this
