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A constant polynomial is that whose value remains the same. It contains no variables. The example for this is P(x) = c. Since there is no exponent so no power to it. Thus, the power of the constant polynomial is Zero. Any constant can be written with a variable with the exponential power of zero. Constant term = 6 Polynomial form P(x)= 6x0
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Gaurav Seth 3 years, 11 months ago
1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________.
(ii) The probability of an event that cannot happen is __________. Such an event is called ________.
(iii) The probability of an event that is certain to happen is _________. Such an event is called _________.
(iv) The sum of the probabilities of all the elementary events of an experiment is __________.
(v) The probability of an event is greater than or equal to and less than or equal to __________.
Solution:
(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
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<a href="https://mycbseguide.com/blog/ncert-solutions-for-class-10-maths-exercise-15-1/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-for-class-10-maths-exercise-15-1/&ved=2ahUKEwiY-tiznffsAhU54zgGHRaTDMcQFjACegQIBBAC" rel="noopener" target="_blank">NCERT Solutions for Class 10 Maths Exercise 15.1 ...</a>
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Gaurav Seth 3 years, 11 months ago
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer:
Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e.,
OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove
ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are
equal.
In ΔAOD and ΔCOD,
OA = OC (Diagonals bisect each other)
∠AOD = ∠COD (Given)
OD = OD (Common)
So, ΔAOD ≅ ΔCOD (By SAS congruence rule)
Hence, AD = CD …………..1
Similarly, it can be proved that
AD = AB and CD = BC ………..2
From equation 1 and 2, we get
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a
parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a
rhombus.
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Gaurav Seth 3 years, 11 months ago
The measure of the angles a, b, c, d are 65°,115°,65°,115°
Step-by-step explanation:
here,
l is parallel to m and p is parallel to q.
then
(vertical opposite angle )
since , l ║ m and q is the transversal line then
(angle on the same side of the transversal )
65 +d = 180°
d = 180-65
d = 115°
now p ║q and m is the transversal line
c +d = 180°(angle on the same side of the transversal )
c + 115° = 180°
c = 180-115
c = 65°
similarly
b +c = 180°
b + 65°= 180
b = 180-65 = 115°
hence ,
The measure of the angles a, b, c, d are 65°,115°,65°,115°
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Gaurav Seth 3 years, 11 months ago
x and y are complementary angles
So, x + y =90 degrees
A. x = 35 degrees
—> x + y = 90 degrees
So, y = 90 - x
hence, y = 90 - 35
Finally, y = 55 degrees
Similarly you can find for B.
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Arjun Chavan 4 years, 1 month ago
1Thank You