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Alexxandr [17]
3 years ago
5

You mix 2 and one half

Mathematics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

wpild be about 3 to 1 juice to pop

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Please help me will mark as brainlist
Nookie1986 [14]

Answer:

17.3825

Step-by-step explanation:

First I figured out the area of the square which is 18×18=324

Then I found out the are of the circle which is 254.47, so 324 - 254.47 = 69.53

Then there are 4 times the shaded are so divided by 4, so 69.53  ÷ 4 = 17.3825

PLEASE GIVE BRAINLIEST!

3 0
3 years ago
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What is the measure of each angle of a regular pentagon? If necessary, round to the nearest tenth.
adelina 88 [10]

Answer: Find the measures of an exterior angle and an interior angle given the number of sides of each regular polygon. Round to the nearest tenth, if necessary. 24 b 15, 345 24, 156 7.5, 172.5 15, 165​

Step-by-step explanation:  sum, S, of the measure of the interior angles of a polygon with n sides is: ... The sum of the interior angles of a 24 -gon is 3960. ... angles of a polygon is (n−2)⋅180 , where n is the number of sides. ... Each exterior angle measures 36024=15 . ... The sum of the 24 interior angles is then 24⋅165=3960 .

7 0
3 years ago
Mikaela purchased a paddleboard originally priced at $699. If all merchandise was on sale for 40% off and the state tax rate is
vitfil [10]

Answer:

299.87

Step-by-step explanation:

699*0.40=279.60\\279.60*0.0725= 20.271 (round)\\279.60+20.27=299.87

6 0
2 years ago
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a landscaping company delivers soul and bags to its customers they deliver 4,000 pounds of soil if each bag holds 16 pounds of s
Murljashka [212]

Answer:

250 bags

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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
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