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Brilliant_brown [7]
3 years ago
5

What is the sum of all integers x that satisfy 1 < (pi -1 )x < 10?

Mathematics
2 answers:
nasty-shy [4]3 years ago
4 0

Answer:

Step-by-step explanation:

pi - 1 = 3.14 - 1 = 2.14

x has to be an integer, and the result must be greater than 1.

1 works.

1 < 2.14*1 < 10  2.14 is between 1 and 10.

2 works

1 < (2.14)*2 < 10

1 < 4.28 < 10

3 works

1 < 2.14 * 3 < 10

1 < 6.42 < 10

4 might work

1 < 2.14*4 < 10

1 < 8.56 < 10

and it does work.

5 can't work

1 < 5 * 2.14 < 10

1 < 10.7 < 10 is not true.

Answer: 1,2,3,4

DedPeter [7]3 years ago
3 0

Answer:

10

Step-by-step explanation:

pi is 3.14...

pi - 1 is 2.14...

question becomes 1<2.14x and 2.14x < 10

1/2.14 < x, 0.46... < x

x < 10/2.14, x < 4.67...

combining

0.46< x < 4.67

x can be 1, 2, 3, 4

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Problem page the longer leg of a right triangle is 1ft longer than the shorter leg. the hypotenuse is 9ft longer than the shorte
LenKa [72]
Hello,

To solve this problem we want to use the Pythagorean Theorem. 
The pythagorean theorem states that for a 90° triangle, 

a^{2} +  b^{2}  =  c^{2}

where a and b represent the two legs of the triangle, and c represents the hypotenuse. 

Let a = the longer leg and b = the shorter leg.
If the longer leg of the triangle is 1 foot longer than the shorter leg, then
a = b +1. 

If the hypotenuse is 9 feet longer than the shorter leg, then c = b + 9.
Using the equations we created, we can plug them into the Pythagorean Theorem to solve for a, b, and c. 

Doing this, we have:
a^{2} +  b^{2} =  c^{2}
(b+1)^{2}  + b =  (b+9)^{2}

Expanding this, we get b^{2} + 2b + 1 +  b^{2}  =  b^{2} + 18b + 81&#10;&#10; 2b^{2} + 2b + 1 =  b^{2} + 18b + 81&#10;&#10; b^{2}  + 1 = 16b + 81&#10;&#10; b^{2} = 16b + 80&#10; &#10;b^{2} - 16b - 80 = 0&#10;&#10;

Solving for b, we get b = 20, and b = -4.
The length of the side of a triangle cannot be negative, so we know that b = 20. 

However, we should check this with the original question to make sure it checks out.

a = b + 1
a = 20 + 1 = 21

c = b + 9
c = 20 + 9 = 29

So, we have a = 21, b = 20, and c = 29. (Also, 20-21-29 is a well known Pythagorean triple)
Using the Pythagorean Theorem, we have:

21^{2} +  20^{2} =  29^{2}
441 + 400 = 841
841 = 841, checks out.

So, the shorter leg is 20 feet, the longer leg is 21 feet, and the hypotenuse is 29 feet. 

Hope this helps!

7 0
2 years ago
How many phone numbers are possible for one area code if the first four numbers are 202-1 , in that order , and the last three n
Mkey [24]

6 phone numbers are possible for one area code if the first four numbers are 202-1

<u>Solution:</u>

Given that, the first four numbers are 202-1, in that order, and the last three numbers are 1-7-8 in any order  

We have to find how many phone numbers are possible for one area code.

The number of way “n” objects can be arranged is given as n!

Then, we have three places which changes, so we can change these 3 places in 3! ways

n! = n \times (n - 1)!

Hence 3! is found as follows:

3! = 3 \times (3-1)!\\\\3! = 3 \times 2!\\\\3! = 3 \times 2 \times 1 = 6

So, we have 6 phone numbers possible for one area code.

3 0
3 years ago
Please help me with this question.
Gelneren [198K]

Answer 1/4

Step-by-step explanation:

3/12 divided by 3 is 1/4

7 0
2 years ago
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victor want to make a book tower of height 48 cm .If the thickness of book is 12 mm ,how many books will he need to make the des
lions [1.4K]

Answer:

40

Step-by-step explanation:

1. Change 48 cm to mm, which will give you 480 mm

2. Divide 480 mm by 12 mm, that will give you 40

Therefore, Victor needs 40 books to make his desired height.

6 0
2 years ago
Garden A has 9 more currant bushes than Garden B. If 3 currant bushes are transplanted from Garden B to Garden A, then Garden A
marusya05 [52]

Answer:

42

Step-by-step explanation:

Let x = the original number of bushes in garden A

Let x-9 = the original number of bushes in garden B

x-9-3=x-12 = bushes in garden B after transplanting 3

x+3 = bushes in garden A after transplanting 3

x+3=1.5(x-12)

x+3=1.5x-18

0.5x=21

x=42

There were 42 current bushes in garden A

3 0
3 years ago
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