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3 years ago

Solve for k. m=−8j−9k+9

Mathematics

2 answers:

Svet_ta [14]3 years ago

8 0

(m + 8j - 9) / -9 ............

inn [45]3 years ago

7 0

Answer: k = -8/9j + -1/9m + 1

Step 1: Flip the equation.−8j − 9k + 9 = m

Step 2: Add -9 to both sides.−8j − 9k + 9 + −9 = m + −9−8j − 9k = m − 9

Step 3: Add 8j to both sides.−8j −9k + 8j = m −9 + 8j−9k = 8j + m −9

Step 4: Divide both sides by -9.
-9k/-9 = 8j + m - 9/-9
k = -8/9j + -1/9m + 1

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Read 2 more answers

A restaurant sells about 330 sandwiches each day at a price of $6 each $.25 decrease in price. 15 more sandwiches are sold per d

zmey [24]

Answer:

restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.

the revenue is $1983.75

Step-by-step explanation:

to calculate current revenue= $6 x 330 = $1980

suppose x as the number of times the price to be dropped by $0.25

then find new price.. i.e

new price= $(6-0.25x)

and, new sell=330 +15x sandwiches

therefore, the new revenue would be= (6-0.25x)(330 +15x)

in order to maximize the current revenue, simplify the above equation and make it complete square using x

(6-0.25x)(330 +15x)

=1980-82.5x +90x -3.75 $x^{2}$

=1980 + 7.5x -3.75 $x^{2}$

=1980-3.75 (-2x+ $x^{2}$ ) ----> taking out common

now, to make a complete square lets add and subtract 1 inside the parentheses

=1980-3.75(-1+1-2x+ $x^{2}$ )

=1980 +3.75 -3.75( $x^{2}$ -2x +1)

=1983.75 -3.75 $(x-1)^{2}$ ---->(1)

as $(x-1)^{2}$ is positive always, minimize the other term in order to maximize the total revenue.

so the minimum possible value of $(x-1)^{2}$ = 0

therefore, x=1

putting x in eq(1) the revenure becomes,

$(1983.75-0)=> $1983.75

therefore, restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.

the revenue is $1983.75

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3 years ago