The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
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Answer:
Each tray has 5 colours and one of the colour on the tray is purple. Hence 1/5 of the colours on the tray is purple and thus 1/5 of the colours on the 20 trays is purple. The fraction of colours on the trays will not change. If written out logically, 20 purples out of 100 colours in total on the trays will also given the fraction 1/5.
Answer:
A
Step-by-step explanation:
the standard form of a quadratic equation is
ax² + bx + c = 0 : a ≠ 0
obtain (5 + x)(5 - x) = 7 in this form by expanding the factors
25 - x ² = 7 ( subtract 7 from both sides )
18 - x² = 0 ( multiply through by - 1 )
x² - 18 = 0 ← in standard form
with a = 1, b= 0 and c = - 18 → A
Answer:
(5(x-4))0
evaluate
1
Step-by-step explanation:
your answer is one i believe
Answer:
y=-4x+36
Step-by-step explanation:
y=mx+b
m is slope or rate; so m = -4
y=-4x + b
10=(-4×4) + b
36 = b
y=-4x+36
