Answer:
3^4 x 3^-7
Step-by-step explanation:
Answer:
Yes
Explaination:
They are similar and the ratio is still the same. The artwork's shape didn't change we just resized it by multiplyting its length and width by 2.
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Given:
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y = - 4x + 16 ;
4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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First, let us simplify the second equation given, to get rid of the "0" ;
→ 4y − x + 4 = 0 ;
Subtract "4" from each side of the equation ;
→ 4y − x + 4 − 4 = 0 − 4 ;
→ 4y − x = -4 ;
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So, we can now rewrite the two (2) equations in the given system:
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y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;
4y − x = -4 ; ===> Refer to this as "Equation 2" ;
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Solve for "x" and "y" ; using "substitution" :
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We are given, as "Equation 1" ;
→ " y = - 4x + 16 " ;
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→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;
to solve for "x" ; as follows:
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Note: "Equation 2" :
→ " 4y − x = - 4 " ;
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Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;
→ as follows:
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→ " 4 (-4x + 16) − x = -4 " ;
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Note the "distributive property" of multiplication :
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a(b + c) = ab + ac ; AND:
a(b − c) = ab <span>− ac .
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As such:
We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
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AND:
→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
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Now, we can write the entire equation:
→ " -16x + 64 − x = - 4 " ;
Note: " - 16x − x = -16x − 1x = -17x " ;
→ " -17x + 64 = - 4 " ; Solve for "x" ;
Subtract "64" from EACH SIDE of the equation:
→ " -17x + 64 − 64 = - 4 − 64 " ;
to get:
→ " -17x = -68 " ;
Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -17x / -17 = -68/ -17 ;
to get:
→ x = 4 ;
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Now, Plug this value for "x" ; into "{Equation 1"} ;
which is: " y = -4x + 16" ; to solve for "y".
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→ y = -4(4) + 16 ;
= -16 + 16 ;
→ y = 0 .
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The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Now, let us check our answers—as directed in this very question itself ;
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→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;
→ Let us check;
→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;
→ Consider the first equation given in our problem, as originally written in the system of equations:
→ " y = - 4x + 16 " ;
→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?
→ "0 = ? -4(4) + 16 " ?? ; → "0 = ? -16 + 16 ?? " ; → Yes! ;
{Actually, that is how we obtained our value for "y" initially.}.
→ Now, let us check the other equation given—as originally written in this very question:
→ " 4y − x + 4 = ?? 0 ??? " ;
→ Let us "plug in" our obtained values into the equation;
{that is: "4" for the "x-value" ; & "0" for the "y-value" ;
→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.
→ " 4(0) − 4 + 4 = ? 0 ?? " ;
→ " 0 − 4 + 4 = ? 0 ?? " ;
→ " - 4 + 4 = ? 0 ?? " ; Yes!
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→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
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→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
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Hope this lenghty explanation is of help! Best wishes!
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Answer: 6x - 3 over x^3
Step-by-step explanation:
Of you have 3 squared you get 9 then 2 times -2 is -4. then you add them 9=-4 which would be ...A: 5
