Given three elements A, B, and C. If compound AB2 is 73.7% by weight of B, and compound BC2 is 28% by weight B, what is the percentage of B in the compound AB2C4?
Answers
Explanation:
The three elements are listed below
Rusting
The process by which iron changes to iron oxide when exposed to oxygen and moisture in the air is called rusting
A = ?
L = 2
B = 1
No SPAMMINGA = ?
L = 2
B = 1
Area = l × b
A = 2 ×1
A = 2❄️ Question :-
\bf \frac{4+3 \sqrt{5} }{4-3 \sqrt{5} } + \frac{4-3 \sqrt{5} }{4+3 \sqrt{5} }4−354+35+4+354−35
Simplify this by rationalizing the denominator.
\begin{gathered} \\ \end{gathered}
❄️ Solution :-
Rationalizing both -
\mapsto \bf \frac{4+3 \sqrt{5} }{4-3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} }↦4−354+35×4+354+35
Using the identities, (a+b)²=a²+b²+2ab and (a-b)(a+b)=a²-b².
\implies \small \bf \dfrac{(4)²+(3 \sqrt{5}) ²+2(4)(3 \sqrt{5} )}{(4)²-(3 \sqrt{5)²} }⟹(4)²−(35)²(4)²+(35)²+2(4)(35)
\implies \bf \frac{16 + 45 + 24 \sqrt{5} }{16 - 45}⟹16−4516+45+245
\implies \bf \frac{61 + 24\sqrt{5} }{ - 29}⟹−2961+245
\mapsto\bf \frac{4 - 3 \sqrt{5} }{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4-3 \sqrt{5} }↦4+354−35×4−354−35
\implies \bf \frac{4 - 3 \sqrt{5} }{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4 - 3 \sqrt{5} }⟹4+354−35×4−354−35
\implies \Large \bf \frac{(4)²+(3 \sqrt{5})²-2(4)(3 \sqrt{5}) }{(4)²-(3 \sqrt{5})² }⟹(4)²−(35)²(4)²+(35)²−2(4)(35)
\implies \bf \frac{16+45 - 24 \sqrt{5} }{(4)² - (3 \sqrt{5})² }⟹(4)²−(35)²16+45−245
\implies \bf \frac{61-24 \sqrt{5} }{16-45}⟹16−4561−245
\implies \bf \frac{61 - 24 \sqrt{5} }{ - 29}⟹−2961−245
Solving them :-
\mapsto \bf \frac{61 + 24 \sqrt{5} }{ - 29} + \frac{61 - 24 \sqrt{5} }{ - 29}↦−2961+245
❄️ Question :-
l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC =~ ∆CDA.
\begin{gathered} \\ \end{gathered}
❄️ Solution :-
✍️Given :- l and m are two parallel lines intersected by p and q.
✍️To prove :- We have to prove that ∆ABC =~ ∆CDA.
✍️Proof :-
In ∆ABC and ∆CDA,
AC = AC [Common side]
<CAB = <ACD [Alternative interior angles]
<ACB = <CAD [Alternative interior angles]
\green \mapsto↦ By ASA congruence rule,
∆ABC =~ ∆CDA.
✍️Hence proved !
\begin{gathered} \\ \end{gathered}
❄️ Question :-
\bf \frac{4+3 \sqrt{5} }{4-3 \sqrt{5} } + \frac{4-3 \sqrt{5} }{4+3 \sqrt{5} }4−354+35+4+354−35
Simplify this by rationalizing the denominator.
\begin{gathered} \\ \end{gathered}
❄️ Solution :-
Rationalizing both -
\mapsto \bf \frac{4+3 \sqrt{5} }{4-3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} }↦4−354+35×4+354+35
Using the identities, (a+b)²=a²+b²+2ab and (a-b)(a+b)=a²-b².
\implies \small \bf \dfrac{(4)²+(3 \sqrt{5}) ²+2(4)(3 \sqrt{5} )}{(4)²-(3 \sqrt{5)²} }⟹(4)²−(35)²(4)²+(35)²+2(4)(35)
\implies \bf \frac{16 + 45 + 24 \sqrt{5} }{16 - 45}⟹16−4516+45+245
\implies \bf \frac{61 + 24\sqrt{5} }{ - 29}⟹−2961+245
\mapsto\bf \frac{4 - 3 \sqrt{5} }{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4-3 \sqrt{5} }↦4+354−35×4−354−35
\implies \bf \frac{4 - 3 \sqrt{5} }{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4 - 3 \sqrt{5} }⟹4+354−35×4−354−35
\implies \Large \bf \frac{(4)²+(3 \sqrt{5})²-2(4)(3 \sqrt{5}) }{(4)²-(3 \sqrt{5})² }⟹(4)²−(35)²(4)²+(35)²−2(4)(35)
\implies \bf \frac{16+45 - 24 \sqrt{5} }{(4)² - (3 \sqrt{5})² }⟹(4)²−(35)²16+45−245
\implies \bf \frac{61-24 \sqrt{5} }{16-45}⟹16−4561−245
\implies \bf \frac{61 - 24 \sqrt{5} }{ - 29}⟹−2961−245
Solving them :-
\mapsto \bf \frac{61 + 24 \sqrt{5} }{ - 29} + \frac{61 - 24 \sqrt{5} }{ - 29}↦−2961+245
a system of government by the whole population or all the eligible members of a state, typically through elected representatives.ᴀʀᴇᴀ ᴏғ ᴛʜᴇ sᴇɢᴍᴇɴᴛ
= \pi \: r^{2} \times \frac {\theta}{360°}πr2×360°θ
= \frac {22}{7} × 7 × 7 × \frac {30°}{360°}722×7×7×360°30°
= 22 × 7 × \frac {1}{12}22×7×121
= 12.83 \: approx.12.83approx.
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