The answer is: [B]: (13, -5) .
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Explanation: Using Substitution:
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y = 8 − x ;
7 = 2 − y ;
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Solve for "x" and "y" using using substitution;
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Take the second equation given:
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7 = 2 − y ;
And considering the first equation given: " y = 8 − x " ;
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Substitute the value: "8 − x" for (the value of "y" in the first equation given), in order to solve for "x" ;
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7 = 2 − y ;
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7 = 2 − 8 − x ;
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7 = -6 − x ;
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Add "6" to EACH side of the equation:
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7 + 6 = -6 − x + 6 ;
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to get: 13 = -x ;
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13 = -x ; ↔ -x = 13 ; ↔ -1 x = 13 ;
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Divide EACH side of the equation by "-1 " ; to isolate "x" on one side of the equation; and to solve for "x" ;
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-1 x / -1 = 13 / -1 ;
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x = 13 . At this point, the only answer choice that can be correct is answer choice: "[B]: (13, -5)" ; since it is the only answer choice given with a solution of x = 13;
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Nonetheless, we shall continue.
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Now that we have the solution for "x"; "x = 13"; We shall plug this value into, or "SUBSTITUTE" this value, "13", for "x", into the first equation given, " y = 8 − x " ; to find the value of "y" ; as follows:
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y = 8 − x ;
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y = 8 − 13 ;
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y = -5 .
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So, we have x = 13; and y = -5; or write as: (13, -5) . This solution corresponds directly which answer choice: [B]:
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Also, to confirm, let us plug in our solved value for "y", which is "-5" ; into our second equation, to see if the equation holds true:
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7 = 2 − y ;
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7 =? 2 − (-5) ?? ;
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7 =? 2 + 5 ? ? Yes !
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Step-by-step explanation:
Numer of events divided by total number of events.
number of events = two ( since T is only 2)
total number of events = 12 ( count all of them)
so the probability is = 2/12 which is = 1/6.
Hope this helps.
Good luck ✅.
Answer:
69 pennies
Step-by-step explanation:
Below is the given data:
Pennies at the starting of February = 34
Pennies in 1st week of February = 13
Pennies in 2nd week of February = 9
Pennies in 3rd week of February = 6
Pennies in 4th week of February = 7
To find the number of pennies at the end of February just add all the pennies.
At the end of February, the number of pennies = 34 + 13 + 9 + 7 +6 = 69 pennies
I can answer individual questions, but i don't know any place where you can get all the answers
