Answer:
1/6 of a quart
Step-by-step explanation:
because
That would be b
hope this helps
![\huge\bold{Given:}](https://tex.z-dn.net/?f=%5Chuge%5Cbold%7BGiven%3A%7D)
Length of the perpendicular "
" = 5.
Length of the hypotenuse "
" = 12.
![\huge\bold{To\:find:}](https://tex.z-dn.net/?f=%5Chuge%5Cbold%7BTo%5C%3Afind%3A%7D)
The length of the missing side ''
".
![\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}](https://tex.z-dn.net/?f=%5Clarge%5Cmathfrak%7B%7B%5Cpmb%7B%5Cunderline%7B%5Corange%7BSolution%7D%7D%7B%5Corange%7B%3A%7D%7D%7D%7D%7D)
The length of the missing side (base) "
" is
.
![\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}](https://tex.z-dn.net/?f=%5Clarge%5Cmathfrak%7B%7B%5Cpmb%7B%5Cunderline%7B%5Cred%7BStep-by-step%5C%3Aexplanation%7D%7D%7B%5Corange%7B%3A%7D%7D%7D%7D%7D)
Using Pythagoras theorem, we have
![({perpendicular})^{2} + ({base})^{2} = ({hypotenuse})^{2} \\ \\⇢ {a}^{2} + {b}^{2} = {c}^{2} \\\\ ⇢ {5}^{2} + {b}^{2} = {12}^{2} \\ \\⇢25 + {b}^{2} = 144 \\ \\⇢ {b}^{2} = 144 - 25 \\ \\⇢ {b}^{2} = 119 \\ \\⇢b = \sqrt{119}](https://tex.z-dn.net/?f=%28%7Bperpendicular%7D%29%5E%7B2%7D%20%20%2B%20%20%28%7Bbase%7D%29%5E%7B2%7D%20%20%3D%20%20%28%7Bhypotenuse%7D%29%5E%7B2%7D%20%20%5C%5C%20%5C%5C%E2%87%A2%20%7Ba%7D%5E%7B2%7D%20%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20%20%7Bc%7D%5E%7B2%7D%20%20%5C%5C%5C%5C%20%E2%87%A2%20%7B5%7D%5E%7B2%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20%20%7B12%7D%5E%7B2%7D%20%20%5C%5C%20%5C%5C%E2%87%A225%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20144%20%5C%5C%20%5C%5C%E2%87%A2%20%7Bb%7D%5E%7B2%7D%20%20%3D%20144%20-%2025%20%5C%5C%20%5C%5C%E2%87%A2%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20119%20%5C%5C%20%5C%5C%E2%87%A2b%20%3D%20%20%5Csqrt%7B119%7D%20)
![\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:side\:"b"\:is\:√119.}](https://tex.z-dn.net/?f=%5Csf%5Cblue%7BTherefore%2C%5C%3Athe%5C%3Alength%5C%3Aof%5C%3Athe%5C%3Amissing%5C%3Aside%5C%3A%22b%22%5C%3Ais%5C%3A%E2%88%9A119.%7D)
![\huge\bold{To\:verify :}](https://tex.z-dn.net/?f=%5Chuge%5Cbold%7BTo%5C%3Averify%20%3A%7D)
![{5}^{2} + { \sqrt{119} }^{2} = {12}^{2} \\\\ ⇝25 + 119 = 144 \\ \\⇝144 = 144 \\ \\⇝L.H.S.=R. H. S](https://tex.z-dn.net/?f=%20%7B5%7D%5E%7B2%7D%20%2B%20%20%7B%20%5Csqrt%7B119%7D%20%7D%5E%7B2%7D%20%20%20%3D%20%20%7B12%7D%5E%7B2%7D%20%20%5C%5C%5C%5C%20%E2%87%9D25%20%2B%20119%20%3D%20144%20%5C%5C%20%5C%5C%E2%87%9D144%20%3D%20144%20%5C%5C%20%5C%5C%E2%87%9DL.H.S.%3DR.%20H.%20S)
Hence verified. ✔
![\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{♡}}}}}](https://tex.z-dn.net/?f=%5Clarge%5Cmathfrak%7B%7B%5Cpmb%7B%5Cunderline%7B%5Corange%7BMystique35%20%7D%7D%7B%5Corange%7B%E2%99%A1%7D%7D%7D%7D%7D)
Answer:
Step-by-step explanation:
The pH of solution a decreases, as the pH of solution B decreases as well.