How do i solve question 25
Answers
I never learned to do operations with absolute values, so I don't have
the standard tools to use. The only thing I have to work with when I run
into one of these is my brain. So there are probably better ways to do it,
but here's how I would approach this one:
The fraction itself is either positive or negative. Either way, its absolute
value is less than 4.
If the fraction is positive, then you can just throw away the absolute-value
bars, and work it like a normal inequality ... multiply each side by (x-3), then
work it down so there's nothing but 'x' on the left side, etc.
If the fraction is negative AND its absolute value is less than 4, then the
fraction itself is greater than -4 . Throw away the absolute-value bars,
change the inequality to say that, and then work it like a normal inequality.
Do BOTH of these, and you'll come out with two inequalities for 'x' ... it's
more than something AND less than something else. Put these together
into one inequality, and you have the whole range for what 'x' can be.
Sounds complicated. One of these days, I should learn the right way
to do these.
2 < x < 14/3
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? cookies in a 12 day period
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different types of coins there are. All we know is that all 10 of them
on the scale at the same time have 45 grams of mass.
IF all the coins happen to be identical ... AND each one is worn down
the same amount, AND each one has the same amount of dirt and oil
on it ... then each one has (45/10) = 4.5 grams of mass.
But those are some big IFs.
There are 4.5 grams in each coin.
equation in ax^2 + bx + c = 0 form. Then find the value of the discriminant to support your
answer.
9. Think of another quadratic equation that has one (1) real number solution. Write the
equation in ax^2 + bx + c = 0 form. Then find the value of the discriminant to support your
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10. Think of another quadratic equation that has no (0) real number solutions. Write the
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Answers
just write 2 factors and expand
(x-3)(x-4)=0
x^2-7x+12=0
discriminant is when you have b^2-4ac
so subsitute
ax^2+bx+c=0
a=1
b=-7
c=12
-7^2-4(1)(12)
49-48=1
discriminant=1
one real solution would be
(x-1)(x-1)=0
x^2-2x+1=0
discriminant=-2^2-4(1)(1)=4-4=0
discriminant=0
no real solutions has the discriminant less than zero
let's say that b=1 and c=20 and a=1
basically, we need a set of numbers where b^2 is less than 4acso
1^2-4(1)(20)=1-80=-79
then we subsitute
x^2+x+20=0
Answers
Answer:
28m you are welcome thank you
Therefore you would do 7x7=49cm2