The answer is B when you subtract every like term
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:
a) 11%
b) 56%
Step-by-step explanation:
a) A= πr^2 6^2π= 36π A= 18^2π = 324π
36π/324π = .111111111111111111111111 or 11%
remember to cancel out the pi symbols when you are showing your work
(put a line across the pi symbols when dividing)
b) A= πr^2 18^2 π = 324π
A= πr^2 12^2π = 144π
324 – 144 = 180
180π/324π = .5555555555555555555555556 or 56%
remember to cancel out the pi symbols when you are showing your work
(put a line across the pi symbols when dividing)
What do you need help with? There is no picture or anything so...
Remember
x^-m=1/(x^m)
u^-3=1/(u^3)
1/(u^3)=1/125
therefor
u^3=125
u=5
x=-3
easy
y=-3-13
y=-16
correct
normally, y is output
(x,y)
chose al second numbers
0,3 are all acceptable
B and D
y=mx+b
m=-3
y=-3x+b
sub point
(x,y)
(1,-2)
-2=-3(1)+b
-2=-3+b
1=b
y=-3x+1
remember (x,y) you keep swtiching those
