The value of x equals 30.
Step by Step explanation:
All sides are equal so you know the angle lengths are equal. When you multiple 2x by 3 you get 6x. All angles of a triangle add up to 180 so you divide 180 by 6 and you get 30. Hope This Helps!
Answer:
Option 4 is correct. The length of PR is 6.4 units.
Step-by-step explanation:
From the given figure it is noticed that the triangle PQR and triangle MQR.
Let the length of PR be x.
Pythagoras formula
Use pythagoras formula for triangle PQM.
The value of PM is 10. The length of PR is x, so the length of MR is (10-x).
Use pythagoras formula for triangle PQR.
.....(1)
Use pythagoras formula for triangle MQR.
.... (2)
From equation (1) and (2) we get
Therefore length of PR is 6.4 units and option 4 is correct.
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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