<em>The</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>parallel</em><em>.</em>
<em>hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>
Divide both sides by -2. We know that since 2 is negative, it'll be < (less than) 0. So, the direction of inequality will change.
Divide 26 by -2..we'll get the answer as -13.
Now, subtract 3 from both the sides.
Subtracting -13 by 3..we'll get the answer as -16.
Divide both sides by 2. We know that since 2 is positive, the direction of inequality will remain the same.
Divide -16 by 2..we'll get the answer as -8.
- The correct answer is ⇨ <u>x </u><u>greater </u><u>than </u><u>or </u><u>equal</u><u> to</u><u> </u><u>-</u><u>8</u><u>.</u>
Its H and K.... they are the only 2 dots that are not in the close group
Step-by-step explanation:
the positive integer numbers that are divisible by 7 are an arithmetic sequence by always adding 7 :
a1 = 7
a2 = a1 + 7 = 7+7 = 14
a3 = a2 + 7 = a1 + 7 + 7 = 7 + 2×7 = 21
...
an = a1 + (n-1)×7 = 7 + (n-1)×7 = n×7
the sum of an arithmetic sequence is
n/2 × (2a1 + (n - 1)×d)
with a1 being the first term (in our case 7).
d being the common difference from term to term (in our case 7).
how many terms (what is n) do we need to add ?
we need to find n, where the sequence reaches 200.
200 = n×7
n = 200/7 = 28.57142857...
so, with n = 29 we would get a number higher than 200.
so, n=28 gives us the last number divisible by 7 that is smaller than 200 (28×7 = 196).
the sum of all positive integers below 200 that are divisible by 7 is then
28/2 × (2×7 + 27×7) = 14 × 29×7 = 2,842
