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3 years ago

8

Compare Partial Products And Regrouping. Describe How The Methods Are Alike And Different.

1 answer:

lisov135 [29]3 years ago

6 0

U have too Cary the number for re gegrouping and I do not know what partial products are

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If you know this answer can you help lol

AnnyKZ [126]

The answer is C whole

4 0

3 years ago

Fred completed 10 math problems. This is 40% of the number of math problems he has to do. How many math problems must he do?

wolverine [178]

10 of the math problems is 40%

40/4 = 10%

10/4 = 2.5 questions

10% = 2.5 questions

10% x 10 = 100%

2.5 questions x 10 = 25 questions

25 questions is the answer

4 0

4 years ago

Read 2 more answers

36 is 80% of what number?

Jlenok [28]

X is the number.
Then:
36=80% of x
80%=80/100=0.8

We can suggest the following equation:
36=0.8x
0.8x=36
x=36/0.8
x=45

Answer: 36 is 80% of 45

5 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-

Charra [1.4K]

Answer:

The statement is true is for any $n\in \mathbb{N}$ .

Step-by-step explanation:

First, we check the identity for $n = 1$ :

$(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}$

$1 = \frac{1\cdot 1\cdot 3}{3}$

$1 = 1$

The statement is true for $n = 1$ .

Then, we have to check that identity is true for $n = k+1$ , under the assumption that $n = k$ is true:

$(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)$

$2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)$

$(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)$

Therefore, the statement is true for any $n\in \mathbb{N}$ .

4 0

3 years ago

You can use a standard caterers symbol, as a chef's hat, on your business card as your

Goryan [66]

It seems like it would make most sense to use a chef's hat on a catering business's business card as a logo.

7 0

3 years ago