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nadya68 [22]
3 years ago
6

AHHH PLEASE HELP ITS MY LAST QUESTION ILL GIVE U FREE ART JUST HALP

Mathematics
1 answer:
ozzi3 years ago
8 0

Answer:

Imao still mad

Step-by-step explanation:

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Prove the following statement.
gayaneshka [121]

Answer:

You can prove this statement as follows:

Step-by-step explanation:

An odd integer is a number of the form 2k+1 where k\in \mathbb{Z}. Consider the following cases.

Case 1. If k is even we have: (2k+1)^{2}=(2(2s)+1)^{2}=(4s+1)^{2}=16s^2+8s+1=8(2s^2+s)+1.

If we denote by m=2s^2+2 we have that (2k+1)^{2}=8m+1.

Case 2. if k is odd we have: (2k+1)^{2}=(2(2s+1)+1)^{2}=(4s+3)^{2}=16s^2+24s+9=16s^{2}+24s+8+1=8(2s^{2}+3s+1)+1.

If we denote by m=2s^{2}+24s+1 we have that (2k+1)^{2}=8m+1

This result says that the remainder when we divide the square of any odd integer by 8 is 1.

6 0
3 years ago
100^a x 1000^b can be written in the form 10^w <br><br> Show that w = 2a + 3b
Elena L [17]
100^a \time 1000^b = 10^{2a} \times 10^{3b} = 10^{2a + 3b}

10^{w} = 10^{2a + 3b}

w = 2a + 3b \text { (shown)}
3 0
3 years ago
A box contains 19 large marbles and 11 small marbles. Each marble is either green or white. 6 of the large marbles are green, an
Anastaziya [24]
9 because if you subtract 11 from 19 you het 9
4 0
3 years ago
Somebody help me solve;
Nadya [2.5K]
P(at least 5 rolls until 1) = P(4 rolls are not 1) = 5/6 x 5/6 x 5/6 x 5/6 = 0.4823 (4sf)

Fewer than 7 rolls to get second 1 after first takes 3 rolls means second occurs on 4th, 5th or 6th roll
The probability of each of these is 1/6, 5/6 x 1/6 and 5/6 x 5/6 x 1/6 respectively.
P(second 1 on 4th, 5th or 6th roll) = 1/6 + 5/36 + 25/216 = 91/216 = 0.4213 (4sf)
3 0
2 years ago
Can someone help me on this question
dexar [7]
It’s an obtuse triangle, this is because none of the sides/angles are equal to one another. Hope this helped you
3 0
3 years ago
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