Pick 2 pairs of equations t<span>hen use addition and subtraction to eliminate </span>the same variable<span> from both pairs of equations then it is left with 2 variables
</span>Pick two pairs
<span><span>4x - 3y + z = - 10</span><span>2x + y + 3z = 0
</span></span>eliminate the same variable from each system
<span><span>4x - 3y + z = - 10</span>
<span>2x + y + 3z = 0</span>
<span>4x - 3y + z = - 10</span>
<span>-4x - 2y - 6z = 0</span>
<span>-5y - 5z = - 10</span>
<span>2x + y + 3z = 0</span>
<span>- x + 2y - 5z = 17</span>
<span>2x + y + 3z = 0</span>
<span>-2x + 4y - 10z = 34</span>
<span>5y - 7z = 34
</span></span>Solve the system of the two new equations:
<span><span>-5y - 5z = - 10</span>
<span>5y - 7z = 34</span>
<span>-12z = 24</span>
which is , <span>z = - 2</span>
<span>-5y - 5(- 2) = - 10</span>
<span>-5y = - 20</span>
wich is , <span>y = 4
</span></span>substitute into one of the original equations
<span>- x + 2y - 5z = 17</span>
<span>- x + 2(4) - 5(- 2) = 17</span>
<span>- x + 18 = 17</span>
<span>- x = - 1</span>
<span>x = 1</span>
<span>which is , </span><span>(x, y, z) = (1, 4, - 2)</span><span>
</span>Does 2(1) + 4 + 3(- 2) = 0<span> ? Yes</span><span>
</span>
<span>Formula for volume of cube: V = a³
</span>V = (5)³ = 5<span> × </span>5<span> × </span><span>5 = 125 cm</span>³
The answer is: [C]: "30%" .
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Explanation:
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Note that: "%" ; or "percent" means "out of 100" ; or "divided by 100" .
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Method 1)
15/50 = ?/ 100 ;
Look at the denominators:
50 * (what value?) = 100 ? ; → "100 ÷ 50 = 2" ;
→ 50 * 2 = 100 ;
So: "15/50 = (15*2)/(50*2) = "30/100" ; which is "30%" .
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Method 2:
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"15/50" = (15÷5) / (50÷5) = 3/10 ;
3/10 = (3*10) / (10*10) = 30/100 ; which is: "30%" .
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Method 3: (slight variation of "Method 2" above):
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"15/50" = (15÷5) / (50÷5) = 3/10 ;
3/10 = 0.3 = 0.30 = (0.30 * 100) % = " 30% " .
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Answer:
12^12
Step-by-step explanation:
can u please help me
What can a denominator never be?
0
so, we need to figure what x can't be... the only way to multiply 2 things together to get 0 is if one or both are zero.
so, what x values make our denominator 0?
to figure this out, we need to set (x-1)(x-2)=0
now we split and solve.

so when x is 1 or 2 the function doesnt make sense.
but, x can be every other number and it does, so the answer is
ALL REAL NUMBERS not equal to 1 or 2