To solve this problem, you'll want to substitute the first equation into the second or the other way around. The equations given are: 1. 3 paperback books + 5 hardcover books = $80.10; 2. 7 paperback books + 4 hardcover books = $100.65. It is helpful to rearrange the first equation to find 5 hardcover books = $80.10 - 3 paperback books, leading to hardcover book = $16.02 - 0.6 paperback books. Now, substitute this into the second equation: 7 paperback books + 4 ($16.02 - 0.6 paperback books) = $100.65, which simplifies to 7 paperback books + $64.08 - 2.4 paperback books = $100.65. This results in 4.6 paperback books = $100.65 - $64.08 = $36.57, thus paperback book = $7.95. You can then use this price in the first equation to determine the hardcover book price: 3 paperback books + 5 hardcover books = $80.10, substituting gives 3($7.95) + 5 hardcover books = $80.10, which leads to 5 hardcover books = $80.10 - $23.85 = $56.25, therefore hardcover book = $11.25. Hence, the total cost for one paperback and one hardcover book is $7.95 + $11.25 = $19.20.
2x^2 - y = -5
x + y = 8
----------------adding gives us
2x^2 + x = 3 <==
** It's important to note that 2x^2 and x cannot be combined, because they are not like terms.
Answer:
The four odd numbers in sequence are 89, 90, 91, and 93.
Step-by-step explanation:
Designate the four consecutive numbers as x, x+2, x+4, and x+6.
Based on the information given in the question
x + (x + 2) + (x + 4) + (x + 6) = 368
4x + 12 = 368
4x = 356
x = 89
Consequently, the numbers are 89, 90, 91, and 93.
Hope this is helpful:)
Response: (0.8115, 0.8645)
Step-by-step outline:
Define p as the proportion of individuals who leave one space after a sentence.
Provided: Sample size: n= 525
Number of respondents indicating they leave one space: 440
Thus, the sample proportion is: 
The z-score for a 90% confidence interval is: 1.645
The formula for determining the confidence interval:


Consequently, a 90% confidence interval for the proportion of people who leave one space after a period is: (0.8115, 0.8645)