Can someone please help me with this?
2 answers:
You might be interested in
Y =ax² + bx +c
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7
Keywords:
<em>System of equations, variables, hardcover version, paperback version, books </em>
For this case we must construct a system of two equations with two variables. Let "h" be the number of hardcover version books, and let "p" be the number of paperback version books. If the hardcover version of a book weighs 7 ounces and the paperback version weighs 5 ounces, to reach a total of 249 ounces we have:
(1)
On the other hand, if there are Forty-five copies of the book then:
(2)
If from (2) we clear the number of books paperback version we have:
As each paperback version book weighs 5 ounces, to obtain the total weight of the paperback version books, represented by "x" in the table shown, we multiply
So,
Answer:
Option D