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4 years ago

9

Ed spent 1 1/4 hours d doing his math homework and 1 3/4 hours doing his science project. Altogether how much time did ed spend

doing homework over the weekend

2 answers:

Setler79 [48]4 years ago

6 0

Altogether Ed spent 3 hours doing his math and science project. 1 1/4 plus 1 3/4 equals 3

Kipish [7]4 years ago

5 0

It will take him 6 hours in total.

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-4x + 64 < 242 + 3x<br> Giving extra points

Zina [86]

Answer:

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Step-by-step explanation:

I'm assuming you wanted to solve for the inequality of x, if so this is the answer.

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3 years ago

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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

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

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.

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4 years ago

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dimulka [17.4K]

The awnser is D but could also maybe be B

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