complaints letter seeking into business purposes
Answers
Answer:
Use this sample letter and these tips to write an effective complaint:
Be clear and concise. ...
State exactly what you want done and how long you're willing to wait for a response. ...
Don't write an angry, sarcastic, or threatening letter. ...
Include copies of relevant documents, like receipts, work orders, and warranties.
Explanation:
Lions are considered as the king of the jungle because they are durable and have a high hunting capacity. The hungry animal has four legs and a tail with healthy paws. Mane is the name of the hair that is on the lion's neck. Lions are carnivorous animals, which is when animals hunt and eat the flesh of other animalsA lion is a wild animal which lives in jungle. He is one of the strongest animals. He is called as “King of the Jungle” due to his huge size, power and attacking nature. Today lions are found in sub-Saharan part of Africa and in Asia. They are represented as a symbol of pride, courage, glory and fearlessness. In ancient times, people, especially kings, used to hunt lions and keep their skin to show their courage and vigour to their enemies.
Lion is a large mammal of cat family. They also have their family with female lion called as “lioness” and offspring called as “cubs”.
We know,
perimeter = 2( l+b)
p=2(2x+x)
p= 2×3x
p= 6x
•°•, Perimeter = 6x(a) the flush point
(b) the flash point
(c) the fire point
(d) all the above
Rusting
The process by which iron changes to iron oxide when exposed to oxygen and moisture in the air is called rusting ❄️ 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 :-
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