does someone mind helping me with this problem? Thank you!

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axel134 axel134 · Answer 1 · 2022-06-14T21:17:14+02:00

[tex]\frac{10-(3 \cdot 2^2 - 23)}{(1 + 10^2)-2^2 \cdot 5^2} =[/tex]

[tex]10-\left(3\cdot \:2^2-23\right) =10-\left(-11\right) =21[/tex]

[tex]\left(1+10^2\right)-2^2\cdot \:5^2 = 101-2^2\cdot \:5^2 = 101-4\cdot \:5^2 =101-4\cdot \:25 =101-100 = 1[/tex]

[tex]=\frac{21}{1}[/tex]

[tex]\frac{a}{1}= a[/tex]

[tex]=21[/tex]

I hope I helped you!

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DᴀʀᴋPᴀʀᴀᴅᴏx DᴀʀᴋPᴀʀᴀᴅᴏx · Answer 2 · 2022-06-15T05:11:05+02:00

[tex]\qquad \sf \dashrightarrow \: \dfrac{10 - (3 \times 2 {}^{2} - 23) }{(1 + 10 {}^{2}) - {2}^{2} \times {5}^{2} } [/tex]

[tex]\qquad \sf \dashrightarrow \: \dfrac{10 - (3 \times 4{}^{} - 23) }{(1 + 10 0{}^{}) - {4}^{} \times {25}^{} } [/tex]

[tex]\qquad \sf \dashrightarrow \: \dfrac{10 - (12- 23) }{(10 1{}^{}) - {100}^{} } [/tex]

[tex]\qquad \sf \dashrightarrow \: \dfrac{10 - ( - 11) }{1} [/tex]

[tex]\qquad \sf \dashrightarrow \: 10 + 11[/tex]

[tex]\qquad \sf \dashrightarrow \: 21[/tex]