Ok so 2(x+1) would be two times the sum of a number and one. “Sum of a number and one” is just “x+1” and then it’s two times that because the two is outside of the parentheses.
And then 2x+1 would be the product of two and a number increased by 1. The product is two and a number is 2x and then increased by 1 is the +1
We are given the price if each ticket, the amount raised, and how many tickets that the STUDENTS sold. So we know that Children tickets cost $3 and adult tickets cost $5, and we also know that they raised $1025. If the students sold 275 tickets, we can assume that these tickets sold by the students where sold TO other students, meaning that we would take 275x3=825 and then use that for 1025-825=200 then use that for 200/5=40 so in total, 40 adult tickets where sold and 275 children tickets where sold. 275+40=315 tickets sold
The solution is when the two functions meet x=2. The green function is f(x), and the red is g(x).
Answer:
D
Step-by-step explanation:
The sides of a rhombus are congruent , then
13b - 9 = 3b + 4 ( subtract 3b from both sides )
10b - 9 = 4 ( add 9 to both sides )
10b = 13 ( divide both sides by 10 )
b = 1.3
The diagonals are perpendicular bisectors of each other , then
14a + 20 = 90 ( subtract 20 from both sides )
14a = 70 ( divide both sides by 14 )
a = 5
Thus c = a + b = 5 + 1.3 = 6.3 → D
Answer: Heyaa!
Your Answer Is... The square root of 121 is 11.
Step-by-step explanation:
<em>It is the positive solution of the equation x2 = 121. The number 121 is a perfect square.</em>
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Prime factorization of 121 = 11 × 11
Prime factors of 11 in pairs: (11 × 11)
Square root of 121: √(11 × 11) = √(112)
Therefore, √121 = 11
Hopefully this helps<em> you !</em>
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- Matthew ~~