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4 years ago

5

Will mark Branliest for whoever answers all 3 right

2 answers:

max2010maxim [7]4 years ago

7 0

A. = 30 000
b. = £3.00
c. = 10 000m

olganol [36]4 years ago

5 0

Answer: a. 30 000

b. £3.00

c.10 000

Step-by-step explanation:

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Write each mixed number as an improper fraction. 1 7/8 1 7/12 2 1/2 NEED BRAINLIEST!!!

Alecsey [184]

15/8 19/12 5/2 is the answer for this problem

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3 years ago

Which transformations of the graph of f(x) = -4x result in the graph of f(x) = 4x +3?

Furkat [3]

Answer:

Option 1

Step-by-step explanation:

If you flip the sign on the coefficient on x, then it will be a reflection over the x-axis. The + 3 is a trick. Make sure you don't forget that you flip the sign to see the direction you are moving the graph horizontally. In this case, it would be moving to the left, so your 1st option is correct.

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3 years ago

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.

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4 years ago

How to round 10.7703 to the nearest tenth.

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