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In-s [12.5K]
3 years ago
14

Students can take MATH 080 between one and four times. The​ student's tuition for MATH 080 is ​$ 548.00 If the tuition is increa

sed 6​% AFTER 2​ ATTEMPTS, what is the​ student's one semester tuition after the 2nd​ attempt? In this situation what is the total tuition for taking MATH 080 the allowed four​ times
Mathematics
1 answer:
melamori03 [73]3 years ago
8 0
2nd attempt carries a price of 6% extra, so (6/100) * 548 is 32.88, so 548 + 32.88, the price is 580.88.

for the 3rd attempt 6% extra of that, (6/100) * 580.88, is 34.8528, so the price is 615.7328.

4th attempt is 6% extra of that, (6/100) * 615.7328, 36.943968, adding that will give us 652.676768.
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Step-by-step explanation:

\underline{ \underline{ \text{Solving \: a \: quadratic \: equation \: by \: factorisation \: method}}} :

In this method , the second order of polynomial ax² + bx + c is factorised and expressed as the product of two linear factors. Then each linear factor is separately solved to get the required solutions of the equation by applying zero factor property. In zero factor property , if p•q = 0 , then either p = 0 or q = 0 . In other words , if the product of two numbers is 0 , then one or both of the numbers must be 0.

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⤑ \tt{ {x}^{2}  + 11x + 5x +  55 = 0}

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\tt{x + 5 = 0 \: \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \: Or :  \:  \:  x + 11 = 0}

\sf{⟶ x  = 0 - 5} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:⟶  x =  0 - 11

\tt{⟶x =  - 5 } \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:     \:  \:⟶   x =  - 11

\red{ \boxed{ \boxed{ { \tt{Our \: final \: answer :  \boxed{ \underline{  \bold{ \tt{x  =  - 5 \: , - 11}}}}}}}}}

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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