Suppose you can factor x^2 + bx + c as (x + q)(x + q). If c>0, what could be possible values of p and q?A. p=5, q=-8
B. p=-2, q=6
C. p=-4, q=-7
D. p=-13, q=1

Answer:

About 43 days

Step-by-step explanation:

Let's assume that the provisions in the hostel are consumed at a constant rate by each student per day. To find out how long the provisions would last with an additional 10 students, we need to consider the total number of students after the new admissions.

Initially, there are 26 students, and the provisions last for 60 days. Therefore, the total provision "student-days" is 26 students multiplied by 60 days, which equals 1560 student-days.

If 10 more students are admitted, the total number of students becomes 26 + 10 = 36 students.

To calculate how many days the provisions would last for 36 students, we divide the total provision "student-days" by the new total number of students:

1560 student-days / 36 students = 43.33 days (approximately)

Therefore, with 10 more students admitted, the provisions would be enough for approximately 43 days.

Answer:

44 days for the 36 students.

Step-by-step explanation:

Let's break down the information given:

Initially, there are 26 students in the hostel and provisions for 60 days. This means that the total "student-days" that the provisions can support is 26 students * 60 days = 1560 student-days.

Now, 10 more students are admitted to the hostel. So, the total number of students becomes 26 + 10 = 36 students.

We want to find out for how many days the provisions will be enough for these 36 students.

We can set up a proportion to solve this:

Initial student-days = New student-days

1560 student-days = 36 students * x days

Now solve for x:

x = 1560 student-days / 36 students

x = 43.33 days

Since you can't have a fraction of a day, we'll round up to the nearest whole day. Therefore, the provisions would be enough for approximately 44 days for the 36 students.

Answer:

(in the image)

Step-by-step explanation:

hope it helps

A geometric sequence is one in which consecutive terms form a fixed ratio r. In other words, if aₙ is the nth term in the sequence, then

For example, if a₁ = a is the 1st term, then

2nd term = a₂ = a₁r

3rd term = a₃ = a₂r = a₁r²

4th term = a₄ = a₃r = a₁r³

and so on. It's fairly easy to infer that

nth term =

14. We're given the 2nd and 5th terms, a₂ = -243 and a₅ = -9, and we use them to find the ratio r.

a₅ = a₄r = a₃r² = a₂r³

-9 = -243 r³

r³ = 1/27

⇒   r = 1/3

Then the 1st term is

a₁ = a₂/r = -243/(1/3) = -729

and the nth term is recursively given by

and explicitly by

15. Now we have a₄ = 1/72 and a₃ = -1/12. Using what we know about geometric sequences, we have

a₄ / a₃ = (a₃r) / a₃ = r

so that

r = (1/72) / (-1/12) = -1/6

Then the 1st term is

a₁ = a₂/r = a₃/r² = (-1/12) / (-1/6)² = -3

and the nth term is recursively given by

and explicitly by