Find the sum of all integers between 84 and 719 which are exactly divisible by 5

50,800

Step-by-step explanation:

The first integer between 84 and 719 that's divisible by 5 is 85.

The last integer between 84 and 719 that's divisible by 5 is 715.

The sum of an arithmetic sequence is:

S = n (a₁ + a) / 2

where n is the number of terms, a₁ is the first term, and a is the last term.

To find the number of terms:

n = (a − a₁) / d + 1

where d is the common difference.

Here, a = 715, a₁ = 85, and d = 5.

n = (715 − 85) / 5 + 1

n = 127

S = 127 (85 + 715) / 2

S = 50,800

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true

1st part is 64,000,000

2nd part 500,000

step-by-step explanation: