Manny wants to find ways to reduce his current debt. The table below shows Manny’s monthly income, expenses and net income.

Monthly Budget
Amount
Income
Wages
$3,000.00
Expenses
Rent
$1,200.00

Utilities
$269.34

Entertainment
$220.00

Food/Clothes
$368.00

Cell Phone
$89.42

Credit Card #1
$94.62

Credit Card #2
$36.18

Car Insurance
$80.00

Car Payment
$300.00

Gym Membership
$25.00
Net Income

$317.44

After evaluating Manny’s monthly budget, which of the following would be the best option to reduce his debt?
a.
Manny could spend less on entertainment each month and put the money towards a new cell phone.
b.
Manny could pay the minimum monthly payments on his credit cards and put the money towards his car payment.
c.
Manny could spend less on food and clothes and put the money towards his credit card payment.
d.
Manny could cut his utlility costs and put the money towards buying a new house.

C

Step-by-step explanation:

Many could spend less on food and clothes and put money towards his credit card payment

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

it has 4

step-by-step explanation:

les

elee

fg

elle

fetish

ser

el

le

f

lent

clee hd

lg

ad

be

hu

te

see

te

ele

cele

her

t

errel

c

s​