I drew it on paper and I got B as the pages that will face each other.
Answer: B. 22 and 23
Given:
Three numbers in an AP, all positive.
Sum is 21.
Sum of squares is 155.
Common difference is positive.
We do not know what x and y stand for. Will just solve for the three numbers in the AP.
Let m=middle number, then since sum=21, m=21/3=7
Let d=common difference.
Sum of squares
(7-d)^2+7^2+(7+d)^2=155
Expand left-hand side
3*7^2-2d^2=155
d^2=(155-147)/2=4
d=+2 or -2
=+2 (common difference is positive)
Therefore the three numbers of the AP are
{7-2,7,7+2}, or
{5,7,9}
Part A is C. because it is $3.49 per mango, so it is $3.49m, and $1.40 per papaya, so that is $1.40p, and $1.24 per coconut, so that is $1.24c. You add those up and the expression would be 3.49m + 1.40p + 1.24c.
Part B is B.
Because 3 times 3.49 is 10.47, which is the cost for the mangoes. And 1.4 times 4 is 5.6, which is the cost for the papayas. And 6 times 1.24 is 7.44, which is the cost of the coconuts. Add 10.47, 5.6, and 7.44 and you get 23.51 which is the answer.
Answer:
m∠P′Q′R′ = m∠PQR
Step-by-step explanation:
For this, we need to find the lowest common multiple of 12 and 15....
common multiples of 12 : 12,24,36,48,60
common multiples of 15 : 15,30,45,60
LCM = 60
the caller that will be the first to win both is the 60th caller