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3 years ago

Find the solution set of the inequality:

Mathematics

2 answers:

NARA [144]3 years ago

7 0

Answer:

-3 + 8 + 15 + 8 - 15 - 3 + 8 = 18

Step-by-step explanation:

I think this is what you meant, I hope this helps <3

8090 [49]3 years ago

5 0

go here for any algebra

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A runner, training for a competition, ran on a track every day for 10 weeks. The first week he ran 8 kilometers each day of the

Levart [38]

Answer:

Total = 686000 meters for 10 weeks 7 days a week.

Step-by-step explanation:

Let's solve this by doing the increase on just 1 training day per week. Then we can find the grand total for 7 days a week by multiplying by 7

a = 8000

n = 10

d = 400

Sum = (a + L)*n/2

L = a + (n - 1)*d
L = 8000 + (10 - 1)*400
L = 8000 + 9*400
L = 8000 + 3600
L = 11600 meters

Sum = (8000 + 11600)*10/2

Sum = 19600 * 10/2

Sum = 98000 meters.

This total represents 1 training day a week.

The total amount is 7 times this figure.

Total = 7 * 98000 = 686 000 meters.

3 0

3 years ago

6x2+40=31x factorise the quadratic equation

user100 [1]

What he said hehehee

8 0

4 years ago

Read 2 more answers

Evaluate 1^3 + 2^3 +3^3 +.......+ n^3

Molodets [167]

Notice that

$(n+1)^4-n^4=4n^3+6n^2+4n+1$

so that

$\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

We have

$\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)$

$\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1$

so that

$\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

You might already know that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

so from these formulas we get

$\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n$

$\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4$

$\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}$

If you don't know the formulas mentioned above:

The first one should be obvious; if you add $n$ copies of 1 together, you end up with $n$ .
The second one is easily derived: If $S=1+2+3+\cdots+n$ , then $S=n+(n-1)+(n-2)+\cdots+1$ , so that $2S=n(n+1)$ or $S=\dfrac{n(n+1)}2$ .
The third can be derived using a similar strategy to the one used here. Consider the expression $(n+1)^3-n^3=3n^2+3n+1$ , and so on.