4(3+14c−12d)
=(4)(3+14c+−12d)
=(4)(3)+(4)(14c)+(4)(−12d)
=12+c−2d
=c−2d+12
The cost of old computer is $ 357
<em><u>Solution:</u></em>
Let "x" be the cost of old computer
From given,
Cost of new computer = $ 709
Blairs new computer cost 5$ less than twice the cost of her old computer
Therefore,
Cost of new computer = twice the cost of old computer - 5
Cost of new computer = 2x - 5
709 = 2x - 5
2x = 709 + 5
2x = 714
Divide both sides by 2
x = 357
Thus the cost of old computer is $ 357
Step One
======
Find the length of FO (see below)
All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)
Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ
Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.
FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??
[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )
Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.
Answer:
-3x^3+74x-5
Step-by-step explanation:
(x+5)(-3x^2+15x-1)
-3x^3+15x^2-x-15x^2+75x-5
-3x^3+15x^2-15x^2+75x-x-5
-3x^3+74x-5
