Find the perimeter of the rectangle whose adjacent sides are 3x-5y-4z and x-3y+5z

Mathematics

1 answer:

Kazeer [188]3 years ago

6 0

<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ I'm assuming the answer is to be left in algebraic form.

Perimeter = length + length + width + width or (2 x length) + (2 x width)

Substitute the values in:

Perimeter = 2(3x - 5y - 4z) + 2(x - 3y + 5z)

Simplify:

Perimeter = 6x - 10y - 8z + 2x - 6y + 10z

Perimeter = 8x - 16y + 2z <== this is the answer

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

You might be interested in

An inchworm (exactly one inch long, of course) is crawling up a yardstick (guess how long that is?). After the rst day, the inch

Snezhnost [94]

3d - 1 , I believe..

8 0

3 years ago

A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min

Natali5045456 [20]

The complete question is;

A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine the amount of salt in the tank at any time

t, if the concentration in the inflow is variable and given by

c(t) = 2 + sin(t/4) lb/gal.

Answer:

dA/dt = (8 + 4sin(t/4)) - (A_t/100)

Step-by-step explanation:

Rate is given as;

dA/dt = R_in - R_out

R_in = (concentration of salt inflow) x (input rate of brine)

So, R_in = (2 + sin(t/4)) x 4 = (8 + 4sin(t/4))

The solution is being pumped out at the same rate, thus it is accumulating at the same rate.

After t minutes, there will be 400 + (0 x t) gallons left = 400 gallons left

Thus,