Alice bought a book for $12. This is 80% of the original price. What was the original

Mathematics

1 answer:

devlian [24]3 years ago

3 0

Answer: $2.40 and he saved $9.60 follows: $12 x . 80 = $9.60 savings

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What are the solutions?<br>8x^2-12x-23=-5

erastova [34]

Answer:

x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

Step-by-step explanation:

Solve for x over the real numbers:

8 x^2 - 12 x - 23 = -5

Divide both sides by 8:

x^2 - (3 x)/2 - 23/8 = -5/8

Add 23/8 to both sides:

x^2 - (3 x)/2 = 9/4

Add 9/16 to both sides:

x^2 - (3 x)/2 + 9/16 = 45/16

Write the left hand side as a square:

(x - 3/4)^2 = 45/16

Take the square root of both sides:

x - 3/4 = (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

x = 3/4 + (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

Answer: x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

4 0

3 years ago

Please solve this question fast.

Alexxandr [17]

Answer:

See Explanation

Step-by-step explanation:

${a} = {6}^{x}, \: \: b = {6}^{y} ...(given) \\ {a}^{y}. {b}^{x} = 36..(1) \\ pluggig \: the \: values \: of \: a \: and \: b \: in \: (1) \\ ({6}^{x}) ^{y} . ({6}^{y}) ^{x} = {6}^{2} \\ {6}^{xy}.{6}^{xy} = {6}^{2} \\ {6}^{xy + xy} = {6}^{2} \\ {6}^{2xy} = {6}^{2} \\ bases \: are \: equal \: so \: exponentes \: will \: also \: be\\ \: equal \\ \therefore \: 2xy = 2 \\ xy = \frac{2}{2} \\ \huge \red { \boxed{ xy = 1}} \\ hence \: proved \\$