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vodka [1.7K]
3 years ago
5

Two equations are given. m + 5n = 20

Mathematics
1 answer:
vesna_86 [32]3 years ago
5 0

Answer:

B. (0, 4)

Step-by-step explanation:

so we need to find the value of m and n, and luckily we have 2 equations that'll help. (I'm going to use the substitution method for this instead of the Elimination method because I am lazy)

first we need to solve for m, but we already have an equation that tells us what m is equal to, m = n - 4.

now we just need to plug in our new m into the other equation and solve for n. (we are using m + 5n = 20 for that)

so now we get

(n - 4) + 5n = 20

so we solve for n

n + 5n = 24

6n = 24

n = 4

now that we know that n is 4, we can plug that into our m equation (m = n - 4)

m = 4 - 4

m = 0

now we have our m and n, and we can make our ordered pair.

(m, n)

(0, 4)

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how many numbers are there that have distinct digits and are greater than 4500? how many 5 digit odd numbers are there that have
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4-digit numbers with distinct digits and greater than 4500: 2800 numbers

5-digit numbers with distinct digits: 27216 numbers.

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If we represent a 4 number digit by ABCD, we have 9 posibilities for A (1,2,3,4,5,6,7,8 and 9, all but 0).

If every digit has to be different, we have 9 posibilities for B: ten digits (0,1,2,3,4,5,6,7,8 and 9 minus the one already used in A).

Int he same way, we have 8 posibilities for C and 7 for D.

Considering all 4-digits numbers, we have 9*9*8*7 = 4536 numers with distinct digits.

To know how many of these numbers are greater than 4500, we can substracte first the numbers that are smaller than 4000: A can take 3 digits (1,2 and 3) and B, C and D the same as before.

3*9*8*7 = 1512 numbers smaller than 4000

Then we can substrat the ones that are between 4000 and 4500

1*4*8*7 = 224 numbers between 4000 and 4500

So, if we substract from the total the numbers that are smaller than 4500 we have the results:

4-digit numbers with distinct digits greater than 4500 = 4536-(1512+224) = 2800

For 5-digit numbers, we can call the number ABCDE.

For A we have 9 digits possible (all but 0).

For B, we also have 9 posibilities (all digits but the one used in A).

For C, we have 8 digits (all 10 but the ones used in A and B).

For D, we have 7 digits.

For E, we have 6 digits.

Multiplying the possible combinations, we have:

9*9*8*7*6 =  27,216 5-digit numbers with distinct digits.

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