Remember that an extraneous solution of an equation is a solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation.
First, we are going to solve our equation algebraically:
Step 1 simplify the equation:
Step 2 subtract 5 from both sides of the equation:
Step 3 square both sides of the equation:
Next, we are going to replace our solution in our original equation and check if it is a valid solution:
Since 9 is not equal to 1, is not valid solution of the equation; therefor it is an extraneous solution.
We can conclude that the correct answer is: x = 16, solution is extraneous
The given system of equations is:
y = 4x − 3 ...(1)
2x + 7y = 41 ...(2)
Notice that the first equation is a solved equation for y. So, we can simply substitute 4x-3 for y in equation (2) to solve. This method is also knwon as substitution method by which we can plug in one equation into other equation.
So, if we will substitute y=4x-3 in equation (2) then we will get,
2x+7(4x-3)=41
Hence, third choice 2x + 7(4x − 3) = 41 is the correct answer.
Work the information to set inequalities that represent each condition or restriction.
2) Name the
variables.
c: number of color copies
b: number of black-and-white copies
3)
Model each restriction:
i) <span>It
takes 3 minutes to print a color copy and 1 minute to print a
black-and-white copy.
</span><span>
</span><span>
3c + b</span><span>
</span><span>
</span><span>ii) He needs to print
at least 6 copies ⇒
c + b ≥ 6</span><span>
</span><span>
</span><span>iv) And must have
the copies completed in
no more than 12 minutes ⇒</span>
3c + b ≤ 12<span />
4) Additional restrictions are
c ≥ 0, and
b ≥ 0 (i.e.
only positive values for the number of each kind of copies are acceptable)
5) This is how you
graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the
first quadrant.
iv) The final region is the
intersection of the above mentioned shaded regions.v) You can see such graph in the attached figure.
We can not answer this without the photo...
A= P(1+r/n)^nt
P<span> =</span> principal amount (the initial amount you borrow or deposit)
r = annual rate of interest (as a decimal)
t<span> = </span>number of years the amount is deposited or borrowed for.
A<span> =</span><span> amount of money accumulated after n years, including interest.</span>
n = number of times the interest is compounded per year
A= 1100(1+0.343/12)^12/4
A = 1197.04
Amount saved = 1197.04 - 1100 = $ 97.04
