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WARRIOR [948]
3 years ago
8

1+3+5+7+...+1003= what is the answer?

Mathematics
2 answers:
Charra [1.4K]3 years ago
4 0
1+(n-1)d = 1003
(n-1)2 = 1002
n-1 = 501
n=502
Sum = 502(1+1003)/2 = 253,004
sergij07 [2.7K]3 years ago
3 0
<span>a) 1 +2 +3 +4 +......+98
98 +97 +96 +95+......+1
sum 99 +99 +99 +99+......+99
there are 98 of these sums, the total is 98 * 99 = 9702, since this sum is twice the sum of the numbers 1 thru 98, we have
sum 1 +2 +3 +4 +.....98 = 9702/2 = 4851

b) 1 +3 +5 +7+.....+1003
1003 +1001 + 999 +997 +....+1
sum 1004 +1004 +1004 +1004 +..1004
There are 502 of these sums. the total is 502 * 1004 = 504008, since this sum is twice the sum of the numbers 1 +3 +5 +7 +....1003, we have
sum 1 +3 +5 +7+.....+1003 = 504008/2 = 252004</span>
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Amanda [17]

Answer:

1) x=-\frac{1}{3}, x=\frac{5}{2}

2) x=1, x=\frac{3}{4}

Step-by-step explanation:

1)

Given

f(y)= 6x^{2} -13x-5=0

Equating above equation.

6x^{2} +2x-15-5=0

2x(3x+1)-5(3x+1)=0

(3x+1)(2x-5)=0

Solving for both the equation

3x+1=0

3x=-1

x=-\frac{1}{3}

2x-5=0

2x=5

x=\frac{5}{2}

x=-\frac{1}{3} ,x=\frac{5}{2}

2)

Given

f(y)=4x^{2} -7x+3=0

4x^{2} -4x-3x+3=0

4x(x-1)-3(x-1)=0

(x-1)(4x-3)=0

Solving for both equation

x-1=0

x=1

4x-3=0

4x=3

x=\frac{3}{4}

x=1, x=\frac{3}{4}

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Step-by-step explanation:


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