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dybincka [34]
4 years ago
10

John wants to send a letter to peter who lives on tesla street john does not remember the house number however he knows that it

has 4 digits it is a multiple of 5 and 7 and that that the last digit is 0 what is the minimum number of letters that john has to send to be sure that peter receives his letter
Mathematics
1 answer:
Serggg [28]4 years ago
5 0

Answer: 137 letters.

Step-by-step explanation:

The data is:

The number has 4 digits.

The number is a multiple of 5

The number is a multiple of 7

The last digit is a 0.

Then, let's find all the possible numbers that met these conditions.

Our number can be formed with 4 digits, A, B, C and D.

D is the last one, so we have  D = 0.

So the number is: ABC0

We know that all the numbers that end with 0 are multiples of 5, so that condition does not matter.

Now, a number is divisible by 7 if:

Take the last digit of the number, double it and subtract it to the remaining number, if the result is multiple of 7, then the initial number is a multiple of 7.

In this case the last digit is 0, so when we double it and subtract it, we actually are not changing the other 3 digits.

Then we have that:

ABC must be a multiple of 7.

So now our problem reduces to find the number of 3-digit numbers that are multiples of 7.

The first multiple of 7 that has 3 digits is:

7*15 = 105.

And the largest possible 3 digit number is 999, now let's try to divide it by 7 to see if this is also a multiple of 7.

999/7 = 142.7

the quotient is not integer, so 999 is not a multiple of 7.

The next option is 998, we can do the same here:

998/7 = 142.6

And so on, we will find that the largest 3 digit multiple of 7 is:

994, such that:

7*152=994.

Then we have:

Smallest: 7*15 = 105

Largest: 7*152 = 994.

Then the number of multiples of 7 between 999 and 100 will be equal to the difference between the quotients between the multples and 7.

105/7 = 15

994/7 = 152

152 - 15 =  137

So we have 137 multiples of 7 with 3 digits.

Then we have 137 4-digit numbers, that end with a zero and also are divisible by 5 and 7.

So John must send 137 letters if he wants to be shure that Peter will get the card.

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A

Step-by-step explanation:

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radian measure = 5 × \frac{\pi }{6} = \frac{5\pi }{6}

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The diameter of a basketball is 9.50 inches. What is the volume of the ball, rounded to the nearest hundredth?
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You have been asked to design a can shaped like right circular cylinder that can hold a volume of 432π-cm3. What dimensions of t
rosijanka [135]

Answer:

Height = 12cm

Radius = 6cm

Step-by-step explanation:

Given

Represent volume with v, height with h and radius with r

V = 432\pi

Required

Determine the values of h and r that uses the least amount of material

Volume is calculated as:

V = \pi r^2h\\

Substitute 432π for V

432\pi = \pi r^2h

Divide through by π

432 = r^2h

Make h the subject:

h = \frac{432}{r^2}

Surface Area (A) of a cylinder is calculated as thus:

A=2\pi rh+2\pi r^2

Substitute \frac{432}{r^2} for h in A=2\pi rh+2\pi r^2

A=2\pi r(\frac{432}{r^2})+2\pi r^2

A=2\pi (\frac{432}{r})+2\pi r^2

Factorize:

A=2\pi (\frac{432}{r} + r^2)

To minimize, we have to differentiate both sides and set A' = 0

A'=2\pi (-\frac{432}{r^2} + 2r)

Set A' = 0

0=2\pi (-\frac{432}{r^2} + 2r)

Divide through by 2\pi

0= -\frac{432}{r^2} + 2r

\frac{432}{r^2} = 2r

Cross Multiply

2r * r^2 = 432

2r^3 = 432

Divide through by 2

r^3 = 216

Take cube roots of both sides

r = \sqrt[3]{216}

r = 6

Recall that:

h = \frac{432}{r^2}

h = \frac{432}{6^2}

h = \frac{432}{36}

h = 12

Hence, the dimension that requires the least amount of material is when

Height = 12cm

Radius = 6cm

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3 years ago
If a test consists of ten multiple choice questions with each permitting a possible answers. How many ways are there in which a
dangina [55]

The question isn't correctly given, a possible format is in the comment below, however, the explanation will cover then concept which can be applied to different but similar

Answer:

10048576 ways

Step-by-step explanation:

We are given a question, from which we can choose aby of 4 options, this gives ua 4 possible choices for 1.

This will also apply if we have more than 1 question with the same number of options.

This can be called the product rule, as each possibility is the same of each question given :

Therefore, given 10 questions:

. We have

4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 = 4^10 = 10048576 ways

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I apologise, but it is as though that there is some missing necessary information. Perhaps there was an error?

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