To simplify this, first, multiply the whole equation with x to get rid of the denominator.
[ x + 3/9/x + 12/6] * x
Simplify 12/6 beforehand, which is 2.
[ x + 3/9/x + 2] * x
Distribute x to each of the terms
x2 + 3/9 + 2x
This is already a simplified equation. If you wan to further simplify by getting rid of the fraction, multiply the whole equation with 9 to eliminate the denominator.
[x2 + 3/9 + 2x]*9
9x2 + 3 + 18x
9x2 + 18x + 3 =0
Label the 3 distinct sides of the box. I arbitrarily chose the letters a, b and c.
Use the info about areas as follows:
ab=54 in^2
ac=90 in^2
bc=60 in^2
Here you have 3 equations in 3 unknowns (a, b and c), which is enough info to use to determine a, b and c. Then the volume of the box is a*b*c.
Example: bc = 90, but c = 60/b. You could subst. 60/b for c in the 2nd and 3rd equation, which will eliminate c completely and leave you with 2 equations in 2 unknowns.
Continuing this procedure, I determined that a=9, b=6 and c=10. Thus, the volume of the box is V = 9*6*10 = 540 cubic inches (answer)
Answer:
Step-by-step explanation:
The cars travel in opposite directions from the same point.
<u>Each hour they travel:</u>
<u>After x hours they will be 425 miles apart:</u>
You take the discount first. Alex is completely right. Laura is if and only if that 110% is really just 10%. So I'll use Alex's method because it is for certain correct.
28.5*35/100 = 9.975 (which is the amount of the discount)
28.5 - 9.975 = 18.525 (which is the amount paid with the discount factored in.)
Now you take 10% of this amount and add it on which gives 18.525 * 10/100=1.825 which is the amount of the tax.
Add this on to the selling price 18.525 + 1.825 = 20.3775 and now is the time to do your rounding.
The cash register will ring up 20.38 which is just under $21.00
Answer: $20.38
Check the picture below.
as you recall from the previous exercise, x = 74°, now, using the "inscribed angle theorem" as you saw already, the green intercepted arc is 160°, so y = 160° - 74°.
as far as ∡z, well, we can just use the "inscribed quadrilateral conjecture".
