Mathematics

Question

Help please!! I will mark brainilist and give a lot of points.




This is Pythagorean theorem.



Formula: a^2+b^2=c^2



When I say ^2 I mean to the power of 2.



The question is asking me to solve for a.



Then give me b and c which is b = 17 1/3 and c = 16.



How do I solve for a when b is greater than c? Won’t that give me a negative a? You can’t have a negative a right?



Unless I’m wrong.




Please help my solve this!!!!! :D

2 Answer

  • Answer:

    what the first guy said

    Step-by-step explanation:

    keep adding by 2 and end up with 22

  • Hello there,

    • I just looked into the question and you are absolutely right whatever you inferred.
    • I think the question might be wrong.
    • The value of c should be greater than b.
    • I guess it will like this:
    • [tex]b = 16 \: \: \: c = 17 \frac{1}{3} [/tex]
    • The values should be like this.
    • I am solving the equation by this:
    • [tex] {a}^{2} = {c}^{2} - {b}^{2} \\ = > {a}^{2} = {(17 \frac{1}{3}) }^{2} - {(16)}^{2} \\ = > {a}^{2} = {( \frac{52}{3} )}^{2} - {(16)}^{2} \\ = > {a}^{2} = ( \frac{52}{3} - 16)( \frac{52}{3} + 16) \\ = > {a}^{2} = ( \frac{52 - 48}{3} )( \frac{52 + 48}{3} ) \\ = > {a}^{2} = \frac{4}{3} \times \frac{100}{3} \\ = > {a}^{2} = \frac{400}{9} \\ = > a = \sqrt{ \frac{400}{9} } \\ = > a = \frac{20}{3} \\ = > a = 6 \frac{2}{3} [/tex]
    • So, the value of a is
    • [tex]a = 6 \frac{2}{3} [/tex]

    Hope you could get an idea from it.

    Doubt clarification - use comment section.

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