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OleMash [197]
3 years ago
10

4 divided by 5/9 = easy question big points lol

Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
8 0

Answer:

7 1/5 or 7.2

Step-by-step explanation:

Thx for points

V125BC [204]3 years ago
5 0
7.2 is answer thanks
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How many proper subsets does the set { 1,2,3} have
OverLord2011 [107]

Answer:

1

Step-by-step explanation:

6 0
2 years ago
Someone please helpppppo
kifflom [539]

Answer:

1: No, they're not similar polygons

2: Yes, they're similar polygons

Step-by-step explanation:

1: DF/AC = 8/6 = 1.3repeating

  FE/CB = 6/4 = 1.5

there is no constant of proportionality

2: 48/40 = 1.2

   35/30 = 1.2

the constant of proportionality is 1.2

3 0
3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
Which statement holds true for a skewed histogram showing a distribution of the weights of students in a class?
Vanyuwa [196]
Among the choices, the statement which describes a skewed histogram showing a distribution of the weights of students in a class is:

<span>"The nature of the skew can be verified by the position of the mean with respect to the mode."

The histogram is skewed to the right if the mean is less than the mode and the histogram is skewed to the left if the mean is more than the mode. </span><span />
8 0
3 years ago
1.
ASHA 777 [7]

Answer:

Step-by-step explanation:

Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,

Rule to be followed,

A(x, y) → P(-y, x)

Therefore, A(1, 1) → P(-1, 1)

Similarly, B(3, 2) → Q(-2, 3)

C(2, 5) → R(-5, 2)

Triangle given in second quadrant will be the triangle PQR.

If the point P of triangle PQR is reflected across a line y = x,

Rule to be followed,

P(x, y) → X(y, x)

P(-1, 1) → X(1, -1)

Similarly, Q(-2, 3) → Y(3, -2)

R(-5, 2) → Z(2, -5)

Therefore, triangle given in fourth quadrant is triangle XYZ.

6 0
3 years ago
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