How many integers between 160 and 279 inclusive, are divisible by 7?

17

Step-by-step explanation:

We can put this as an AP (arithmetic progression).

The first number, after 160, that is divisible by 7 is:

161

The last number before 279 that is divisible by 7 is:

273

We can use the formula:

Where

tn is the nth term [here we take the last one, 273

a is the first term [161]

d is the common difference [7]

and n is the number of terms, we are trying to find this.

Substituting we get:

Hence, there are 17 integers

need more info

step-by-step explanation:

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1