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3 years ago

Solve for a. 7a-2b = 5a+b

Mathematics

2 answers:

andrew11 [14]3 years ago

8 0

Answer: The required value of a is $\dfrac{3}{2}b.$

Step-by-step explanation: We are given to solve the following equation for variable a :

$7a-2b=5a+b~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

To solve the above equation for a, we need to take terms involving a on one side of equality that that involving b on the other side.

So, from equation (i), we get

$7a-2b=5a+b\\\\\Rightarrow 7a-5a=b+2b\\\\\Rightarrow 2a=3b\\\\\Rightarrow a=\dfrac{3}{2}b.$

Thus, the required value of a is $\dfrac{3}{2}b.$

nordsb [41]3 years ago

7 0

7a-2b = 5a + b
2a=3b
a = 3b/2

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4 years ago

how many numbers are there that have distinct digits and are greater than 4500? how many 5 digit odd numbers are there that have

marusya05 [52]

Answer:

4-digit numbers with distinct digits and greater than 4500: 2800 numbers

5-digit numbers with distinct digits: 27216 numbers.

Step-by-step explanation:

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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3 years ago