The digit in the tens place of a two digit number is three times that in the units place. If the digits are reversed the new num
ber will be 36 less than the original number. Find the original number with one variable
1 answer:
We could do it with algebra. But we can also do it the long way, which is
shorter than the algebraic way.
The digit in the tens place is 3 times the digit in the units place.
So the number MUST be
31, or
62, or
93 .
It can't be anything else.
Now here they are again, with the reverse of each one:
31 . . . 13 The new number is 18 less.
62 . . . 26 The new number is 36 less.
93 . . . 39 The new number is 54 less.
Obviously, the original number is 62.
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Expand first:-
yx^2 - xy^2 + zy^2 -yz^2 + xz^2 - zx^2
there is no common factor so we cant factorize this
Add 4 to both sides
-10 + 4 = 2a
Simplify -10 + 4 to -6
-6 = 2a
Divide both sides by 2
-6/2 = a
Simplify 6/2 to 3
-3 = a
Switch sides
<u>a = -3</u>
Answer:
B
Step-by-step explanation:
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Step-by-step explanation: