<u>Solution-</u>
Given that,
In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.
Then considering ΔPQT and ΔSTF,
1- ∠FTS ≅ ∠PTQ ( ∵ These two are vertical angles)
2- ∠TFS ≅ ∠TPQ ( ∵ These two are alternate interior angles)
3- ∠TSF ≅ ∠TQP ( ∵ These two are also alternate interior angles)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.
As QS = TQ + TS = 10 (given)
If TS is x, then TQ will be 10-x. Then putting these values in the equation
∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm
Answer:
Yes
Step-by-step explanation:
Since we are adding two polynomials
The sum will also be polynomial
Answer:
Step-by-step explanation:
a1=2/3
sequence is 2/3,3/4,4/5,...
for numerator a1=2
d=3-2=1
numerator of nth term=a1+(n-1)d=2+(n-1)×1=2+n-1=n+1
denominator = 1 more than numerator=n+1+1=n+2
so an=(n+1)/(n+2)
or for denominator a1=3,d=4-3=1
denominator of nth term=3+(n-1)×1=3+n-1=n+2
an=(n+1)/(n+2)
Answer:
for all x in the domain of f(x), or odd if, f(−x) = −x, for all x in the domain of f(x), or neither even nor odd if neither of the above are true statements. A kth degree polynomial, p(x), is said to have even degree if k is an even number and odd degree if k is an odd number
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given:
Solving for :