Can someone help and explain and do the solution pls :(

1 answer:

6 0

Consider ∆JWZ and ∆JKZ

WZ~KJ (given)

/ WZJ~/ KJZ (given)

JZ~JZ (common)

Therefore,

∆JWZ~∆JKZ by SAS congruence rule.

JW~ZK by CPCT.

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navik [9.2K]

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Step-by-step explanation:

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Answer:

$C. x$

Step-by-step explanation:

Option A $[x+5=20]$ tells us that: When we add 5 to a variable x, we get 20. As it has a unique value for x and is completely equal to it(i.e. 15), It is an equality.

Option B $[x=5]$ tells us that: A variable x equals to 5. Hence, as x is unique for 5 and is wholly equal to it, it's an equality too.
Option C $[x$ tells us that: A variable x isn't 5 but lesser than it. As we cannot equate it to 5, nor we are given the nature of the variable x, it is an Inequality.
Option D $[x+5]$ is an expression; It can't be called an equation or an inequality unless we relate it with another expression.