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Answer:
The number is 49
Step-by-step explanation:
Let [tex]a[/tex] be the tens digit of the two-digit number. Since the units digit is one more than twice the tens digit ([tex]a[/tex]), the units digit can be represented in terms of [tex]a[/tex] as [tex]2a+1.[/tex] The problem has also given that the sum of the digits is [tex]13[/tex] so we create the equation
[tex]a+(2a+1)=13[/tex]
Solve for [tex]a.[/tex]
[tex]a+2a+1=13[/tex]
[tex]3a+1=13[/tex]
[tex]3a+1-1=13-1[/tex]
[tex]3a=12[/tex]
[tex]\therefore a=\dfrac{12}{3}=4[/tex]
So, the tens digit is [tex]4.[/tex] Now, substitute [tex]a=4[/tex] to [tex]2a+1[/tex] to find the units digit.
[tex]\textsf{units digit}=2a+1=2(4)+1[/tex]
[tex]\textsf{units digit}=8+1[/tex]
[tex]\textsf{units digit}=9[/tex]
By using the results, we can conclude that the two-digit number is [tex]\boxed{49}.[/tex]
Answer:
49
Step-by-step explanation:
given the sum of the digits of a two-digit number is 13. if the number is four more than five times the units digit
to find what is the number.
solution
let the two digits be x and y such that the number is, N= 10x+y
sum of the digits =13
x+y= 13
the number is four more than five times the unit utilized digit.
N=4+5y
10x+y= 4+5y
10x - 4y= 4
5x - 2y=2
using y=13-x from eqn. (1),
5x-2 (13-x)=2
x=4
y=13-4=9
therefore, N=10(4) + 9 =49
i hope it helps:)