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Sagot :
We have two equations.
Let M be the Michaela's age and J the Jake's age.
The first one is that Michaela's age is ten years more than half the age of Jake.
[tex]M=\frac{J}{2}+10[/tex]The second equation can be written knowing that the sum of the two ages is 55:
[tex]M+J=55[/tex]We can start by substituting J in the first equation, using the information of the second equation:
[tex]\begin{gathered} J=55-M \\ M=\frac{J}{2}+10=\frac{(55-M)}{2}+10=\frac{55}{2}-\frac{M}{2}+10=\frac{75}{2}-\frac{M}{2} \\ M=\frac{75}{2}-\frac{M}{2} \\ M+\frac{M}{2}=\frac{75}{2} \\ \frac{3}{2}M=\frac{75}{2} \\ 3M=75 \\ M=\frac{75}{3} \\ M=25 \end{gathered}[/tex]The age of Michaela is 25 years.
7) Lets call
H: the number of atoms of hydrogen
C: the number of atoms of carbon
O: number of atoms of oxygen
We know that there are twice as many atoms of hydrogen as oxygen. This means that if we have 2 oxygens, we should have 4 hydrogens.
This can be written as:
[tex]H=2O[/tex]We know that one more atom of carbon than hydrogen.
This is:
[tex]C=H+1[/tex]We know that the sum of atoms is 21. This is:
[tex]H+C+O=21[/tex]Now that we have 3 equations for the three unknowns we start by solving for C.
We use the substitution method to solve it.
[tex]C=H+1\longrightarrow H=C-1[/tex][tex]H=2O=C-1\longrightarrow O=\frac{C}{2}-\frac{1}{2}[/tex]Then, we can replace H and O in the 3rd equation:
[tex]\begin{gathered} H+C+O=21 \\ (C-1)+C+(\frac{C}{2}-\frac{1}{2})=21 \\ (1+1+\frac{1}{2})C-1-\frac{1}{2}=21 \\ \frac{5}{2}C-\frac{3}{2}=21 \\ \frac{5}{2}C=21+\frac{3}{2}=\frac{42+3}{2}=\frac{45}{2} \\ \frac{5C}{2}=\frac{45}{2} \\ C=\frac{45}{2}\cdot\frac{2}{5}=9 \end{gathered}[/tex]The numbers of atoms of carbon is 9 atoms.
Alternative method:
Atoms of hydrogen = 2 * atoms of oxygen
Atoms of hydorgen = atoms of carbon - 1
Then we can say that:
Atoms of hydrogen = 2 * atoms of oxygen = atoms of carbon -1
Then, from the last equation:
Atoms of oxygen = (1/2)*atoms of carbon - (1/2)
If we add all the atoms, we get 21 atoms:
Atoms of hydrogen + Atoms of oxygen +Atoms of carbon = 21
We will replace each of the types of atoms for an expression where the atoms of carbon are present, because is what we need for our answer.
We will call x the number of atoms of carbon.
As:
Atoms of hydrogen = atoms of carbon - 1 = x-1
Atoms of oxygen = (1/2)*atoms of carbon - (1/2) = 1/2*x-1/2
We will have:
[tex]\begin{gathered} Atomsofhydrogen+Atomsofoxygen+Atomsofcarbon=21 \\ (x-1)+(\frac{1}{2}x-\frac{1}{2})+x=21 \\ x-1+\frac{1}{2}x-\frac{1}{2}+x=21 \\ (x+\frac{1}{2}x+x)-1-\frac{1}{2}=21 \\ \frac{5}{2}x-\frac{3}{2}=21 \\ \frac{5}{2}x=21+\frac{3}{2}=\frac{21\cdot2+3}{2}=\frac{42+3}{2}=\frac{45}{2} \\ x=\frac{45}{2}\cdot\frac{2}{5}=9 \end{gathered}[/tex]x, the number of atoms of carbon, is equal to 9.
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