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Sagot :
Sure, let's solve this step-by-step.
1. Define the Digits:
Let [tex]\( x \)[/tex] be the digit at the ten's place.
The digit at the unit place will then be [tex]\( x + 4 \)[/tex] (since it's stated to be 4 more than the ten's place digit).
2. Set Up the Equation:
According to the problem, the product of these two digits is 21.
Therefore, we have the equation:
[tex]\[ x \times (x + 4) = 21 \][/tex]
3. Solve the Equation:
Expand the equation:
[tex]\[ x(x + 4) = 21 \][/tex]
[tex]\[ x^2 + 4x = 21 \][/tex]
Rearrange to form a standard quadratic equation:
[tex]\[ x^2 + 4x - 21 = 0 \][/tex]
4. Factorize the Quadratic Equation:
Now we need to solve the quadratic equation [tex]\( x^2 + 4x - 21 = 0 \)[/tex].
The solutions to this equation are given by the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex]:
[tex]\[ x = \frac{{-4 \pm \sqrt{{4^2 - 4 \times 1 \times (-21)}}}}{2 \times 1} \][/tex]
[tex]\[ x = \frac{{-4 \pm \sqrt{{16 + 84}}}}{2} \][/tex]
[tex]\[ x = \frac{{-4 \pm \sqrt{100}}}{2} \][/tex]
[tex]\[ x = \frac{{-4 \pm 10}}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{{6}}{2} = 3 \][/tex]
or
[tex]\[ x = \frac{{-14}}{2} = -7 \][/tex]
Since [tex]\( x \)[/tex] represents a digit, it must be a non-negative integer (0-9). Therefore, the valid solution is:
[tex]\[ x = 3 \][/tex]
5. Determine the Two-Digit Number:
If [tex]\( x = 3 \)[/tex], then the digit at the unit place is:
[tex]\[ x + 4 = 3 + 4 = 7 \][/tex]
So, the two-digit number is:
[tex]\[ 37 \][/tex]
Conclusion:
(a) The digit at the ten's place is [tex]\( \boxed{3} \)[/tex].
(b) The natural number is [tex]\( \boxed{37} \)[/tex].
(c) Since we successfully found a number where both conditions (unit digit 4 more than the ten's place digit, product of digits equals 21) are satisfied, the answer is [tex]\( \boxed{\text{Yes}} \)[/tex].
(d) The other parts of your question may be based on specific context not visible in your provided text, so I won't speculate further.
1. Define the Digits:
Let [tex]\( x \)[/tex] be the digit at the ten's place.
The digit at the unit place will then be [tex]\( x + 4 \)[/tex] (since it's stated to be 4 more than the ten's place digit).
2. Set Up the Equation:
According to the problem, the product of these two digits is 21.
Therefore, we have the equation:
[tex]\[ x \times (x + 4) = 21 \][/tex]
3. Solve the Equation:
Expand the equation:
[tex]\[ x(x + 4) = 21 \][/tex]
[tex]\[ x^2 + 4x = 21 \][/tex]
Rearrange to form a standard quadratic equation:
[tex]\[ x^2 + 4x - 21 = 0 \][/tex]
4. Factorize the Quadratic Equation:
Now we need to solve the quadratic equation [tex]\( x^2 + 4x - 21 = 0 \)[/tex].
The solutions to this equation are given by the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex]:
[tex]\[ x = \frac{{-4 \pm \sqrt{{4^2 - 4 \times 1 \times (-21)}}}}{2 \times 1} \][/tex]
[tex]\[ x = \frac{{-4 \pm \sqrt{{16 + 84}}}}{2} \][/tex]
[tex]\[ x = \frac{{-4 \pm \sqrt{100}}}{2} \][/tex]
[tex]\[ x = \frac{{-4 \pm 10}}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{{6}}{2} = 3 \][/tex]
or
[tex]\[ x = \frac{{-14}}{2} = -7 \][/tex]
Since [tex]\( x \)[/tex] represents a digit, it must be a non-negative integer (0-9). Therefore, the valid solution is:
[tex]\[ x = 3 \][/tex]
5. Determine the Two-Digit Number:
If [tex]\( x = 3 \)[/tex], then the digit at the unit place is:
[tex]\[ x + 4 = 3 + 4 = 7 \][/tex]
So, the two-digit number is:
[tex]\[ 37 \][/tex]
Conclusion:
(a) The digit at the ten's place is [tex]\( \boxed{3} \)[/tex].
(b) The natural number is [tex]\( \boxed{37} \)[/tex].
(c) Since we successfully found a number where both conditions (unit digit 4 more than the ten's place digit, product of digits equals 21) are satisfied, the answer is [tex]\( \boxed{\text{Yes}} \)[/tex].
(d) The other parts of your question may be based on specific context not visible in your provided text, so I won't speculate further.
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