Answer:
<h2>42 and -16</h2>
Step-by-step explanation:
![m,\ n-the\ numbers\\\\\text{The equations:}\\\\(1)\qquad n+2m=10\\(2)\qquad n+m=26\\\\\text{subtract both sides of the equations}\\\\(n+2m)-(n+m)=10-26\\\\n+2m-n-m=-16\qquad\text{combine like terms}\\\\(n-n)+(2m-m)+-16\\\\\boxed{m=-16}\\\\\text{Put the value of m to (2)}:\\\\n+(-16)=26\\\\n-16=26\qquad\text{add 16 to both sides}\\\\\boxed{n=42}](https://tex.z-dn.net/?f=m%2C%5C%20n-the%5C%20numbers%5C%5C%5C%5C%5Ctext%7BThe%20equations%3A%7D%5C%5C%5C%5C%281%29%5Cqquad%20n%2B2m%3D10%5C%5C%282%29%5Cqquad%20n%2Bm%3D26%5C%5C%5C%5C%5Ctext%7Bsubtract%20both%20sides%20of%20the%20equations%7D%5C%5C%5C%5C%28n%2B2m%29-%28n%2Bm%29%3D10-26%5C%5C%5C%5Cn%2B2m-n-m%3D-16%5Cqquad%5Ctext%7Bcombine%20like%20terms%7D%5C%5C%5C%5C%28n-n%29%2B%282m-m%29%2B-16%5C%5C%5C%5C%5Cboxed%7Bm%3D-16%7D%5C%5C%5C%5C%5Ctext%7BPut%20the%20value%20of%20m%20to%20%282%29%7D%3A%5C%5C%5C%5Cn%2B%28-16%29%3D26%5C%5C%5C%5Cn-16%3D26%5Cqquad%5Ctext%7Badd%2016%20to%20both%20sides%7D%5C%5C%5C%5C%5Cboxed%7Bn%3D42%7D)
The numbers are: "3" and "9" .
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Explanation:
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Let "x" represent one of the two (2) numbers.
Let "y" represent the other one of the two (2) numbers.
x = 2y + 3 ;
x + y = 12 .
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Method 1)
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x = 12 <span>− y ;
Plug this into "x" for "2y + 3 = x" ;
</span>→ 2y + 3 = 12 <span>− y ;
</span>
Add "y" to each side of the equation; & subtract "3" from each side of the equation ;
→ 2y + 3 + y − 3 = 12 − y + y <span>− 3 ;
</span>
to get: 3y = 9 ;
Divide each side of the equation by "3" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
3y / 3 = 9 / 3 ;
y = 3 .
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Now: x = 12 − y ; Plug in "3" for "y" ; to solve for "x" ;
→ x = 12 − 3 = 9
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So; x = 9, y = 3 .
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Method 2)
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When we have:
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x = 2y + 3 ;
x + y = 12 .
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→ y = 12 − x ;
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Substitute "(12−x)" for "y" in the equation:
" x = 2y + 3 " ;
→ x = 2(12 − x) + 3 ;
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Note the "distributive property of multiplication" :
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a(b + c) = ab + ac ;
a(b − c) = ab − ac ;
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As such:
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→ 2(12 − x) = 2(12) − 2(x) = 24 − 2x ;
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So; rewrite:
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x = 2(12 − x) + 3 ;
as:
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→ x = 24 − 2x + 3 ;
→ x = 27 − 2x ;
Add "2x" to each side of the equation:
→ x + 2x = 27 − 2x + 2x ;
→ 3x = 27 ;
Divide each side of the equation by "3" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
3x / 3 = 27 / 3 ;
x = 9 .
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Note: "y = 12 − x" ; Substitute "9" for "x" ; to solve for "y" ;
→ y = 12 − 9 = 3 ;
→ y = 3 .
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So, x = 9 ; and y = 3.
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The numbers are: "3" and "9" .
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To check our answers:
Let us plug these numbers into the original equations;
to see if the equations hold true ; (i.e. when, "x = 9" ; and "y = 3"
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→ x + y = 12 ;
→ 9 + 3 =? 12 ? Yes!
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→ x = 2y + 3 ;
→ 9 =? 2(3) + 3 ?? ;
→ 9 =? 6 + 3 ? Yes!!
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The graph's function, which depicts an exponential function, needs a domain of x ≥ 1. It's best to choose C.
We must choose which of the three functions in the graph of a function that has three distinct functions for its domain represents an exponential function.
An exponential function is what?
The domain of this exponential function is defined by the formula
f(x) = y = a∧x where a is a constant and x is a variable.
Since the graph continuously drops to -4 as shown throughout, it then displays a parabolic function from -4 to 0.9 before beginning to increase exponentially.
Consequently, the x domain is necessary for the function shown on the graph, which represents the exponential function. The ideal response is C.
Learn more about functions such as exponential , piecewise etc here
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10 procent
Step-by-step explanation:
150 / 150 = 10