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Bezzdna [24]
3 years ago
8

2) A

Mathematics
1 answer:
Natali5045456 [20]3 years ago
6 0
He should take $20 an hour because he will get $50 more than the 15% on all his sales
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Colin has 18 pencils. How many kids (including Colin) can share these pencils evenly?
julsineya [31]
18 kids right? Or am I dumb? I feel like I’m wrong because this question seems so easy.
5 0
3 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.
7 0
4 years ago
The height of a triangle is 3cm less than twice the base. The area is 10cm2
Kamila [148]

Answer:

The base is 4cm and the height is 5cm.

Step-by-step explanation:

This is a solve the system question. Call H the height of the triangle and B the base. The question tells us:

H=2B-3

and

\frac{B \times H}{2}=10

Sub the first equation into the second (as H is already isolated). You will end up with a quadratic equation - solve that any way you wish (e.g. quadratic formula). I've provided the factored form below which shows you the roots:

\frac{B(2B-3)}{2}=10\\\frac{2B^2-3B}{2}=10\\2B^2-3B=20\\2B^2-3B-20=0\\(2B+5)(B-4)=0\\B= - \frac{5}{2},4

In this question, we take B=4. You can't have a negative side length so the other answer is eliminated. Now sub the value for B into either of the original equations. I'll use the first, again because H is already isolated:

H=2(4)-3\\H=8-3\\H=5

3 0
3 years ago
Solve for x... please help!​
d1i1m1o1n [39]

Answer:

x = 69

Step-by-step explanation:

8 0
4 years ago
Plss help it’s a test and it’s due in 10 minutes plsss helpp
I am Lyosha [343]

Answer: Commutative

Step-by-step explanation:

5 0
3 years ago
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