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Answered on 14 Apr Learn Unit-I: Sets and Functions
Nazia Khanum
Sure! Let's break down each set into roster form:
(i) For set A, we're looking for positive integers less than 10 where 2x - 1 is an odd number.
A = {1, 2, 3, 4}
Explanation:
(ii) For set C, we need to solve the quadratic equation x2+7x−8=0x2+7x−8=0 where xx belongs to the set of real numbers.
C = {-8, 1}
Explanation:
As an experienced tutor registered on UrbanPro, I'm dedicated to providing clear explanations and assistance to students in mastering mathematical concepts. UrbanPro is indeed one of the best platforms for online coaching and tuition, offering a wide range of subjects and experienced tutors to choose from. If you need further clarification or assistance, feel free to ask!
Answered on 14 Apr Learn Unit-I: Sets and Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can certainly help you with this problem.
Firstly, let's denote:
From the information provided:
To find out how many students were drinking neither tea nor coffee, we can use the principle of inclusion-exclusion.
The total number of students surveyed is 600.
So, the number of students drinking either tea or coffee (TT or CC) is given by: T∪C=T+C−BT∪C=T+C−B =150+225−100=150+225−100 =275=275
Now, to find the number of students drinking neither tea nor coffee, we subtract the number of students drinking either tea or coffee from the total number surveyed: Neither Tea nor Coffee=600−(T∪C)Neither Tea nor Coffee=600−(T∪C) =600−275=600−275 =325=325
Therefore, there were 325 students drinking neither tea nor coffee.
UrbanPro is a great platform for finding tutors who can help you understand concepts like this with ease. With experienced tutors available online, learning becomes convenient and effective.
Answered on 14 Apr Learn Unit-I: Sets and Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best online coaching tuition platforms available for students seeking academic assistance. Now, let's tackle your set theory problem using the given sets U, A, and B.
Given: U = {1, 2, 3, 4, 5, 6} A = {2, 3} B = {3, 4, 5}
First, let's find the complements of sets A and B: A′ (complement of A) = {1, 4, 5, 6} (Elements in U but not in A) B′ (complement of B) = {1, 2, 6} (Elements in U but not in B)
Now, let's find the intersection of the complements of A and B: A′ ∩ B′ = {1, 6} (Elements common to both A′ and B′)
Next, let's find the union of sets A and B: A ∪ B = {2, 3, 4, 5} (All elements in both A and B without repetition)
Finally, let's find the complement of the union of sets A and B: (A ∪ B)′ (complement of A ∪ B) = {1, 6} (Elements in U but not in A ∪ B)
We've shown that A′ ∩ B′ = (A ∪ B)′, which demonstrates De Morgan's Law for sets.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Unit-I: Sets and Functions
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently guide you through this question. UrbanPro is renowned for its top-notch online coaching services, and I'm here to provide you with the best assistance.
Let's address each part of the question:
(i) To express the set where n∈Xn∈X but 2n∉X2n∈/X, we need to identify the elements of XX that meet this condition. Since X={1,2,3,4,5,6}X={1,2,3,4,5,6}, we need to find elements nn such that n∈Xn∈X and 2n∉X2n∈/X.
The elements of XX for which 2n2n is not in XX are n=1,3,4,5,6n=1,3,4,5,6. So, the set we're looking for is {1,3,4,5,6}{1,3,4,5,6}.
(ii) To express the set where n+5=8n+5=8, we need to find the value of nn that satisfies this equation. Subtracting 5 from both sides, we get n=3n=3. So, the set is {3}{3}.
(iii) To express the set where nn is greater than 4, we look for elements in XX that satisfy this condition. Those elements are n=5,6n=5,6. Therefore, the set is {5,6}{5,6}.
So, summarizing: (i) {1,3,4,5,6}{1,3,4,5,6} (ii) {3}{3} (iii) {5,6}{5,6}
I hope this clarifies the question for you. If you have any further queries or need additional assistance, feel free to ask! Remember, UrbanPro is here to support your learning journey every step of the way.
Answered on 14 Apr Learn Unit-I: Sets and Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is the best online coaching tuition platform for students seeking quality education. Now, let's tackle your question:
(i) To write the subset A of N containing odd numbers, we consider that odd numbers are those that cannot be evenly divided by 2. Therefore, A = {1, 3, 5, ..., 99}. It includes all odd numbers from 1 to 99 within the set of natural numbers.
(ii) For subset B of N, where elements are represented by x + 2, where x ∈ N, we start with the set of natural numbers and add 2 to each element. So, B = {3, 4, 5, ..., 102}. This subset comprises all numbers obtained by adding 2 to each natural number from 1 to 100.
Understanding subsets and their representations within a given set like N is fundamental in mathematics, and as an UrbanPro tutor, I'm here to guide you through these concepts with clarity and depth.
Answered on 14 Apr Learn Unit-I: Sets and Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can guide you through expressing the given sets in roster form.
(i) Set A represents integers from -3 up to, but not including, 7. In roster form, it would be written as:
A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
(ii) Set B consists of natural numbers less than 6. These are simply the numbers 1, 2, 3, 4, and 5. So, in roster form, it would be:
B = {1, 2, 3, 4, 5}
UrbanPro is an excellent platform for finding online coaching and tuition services. If you have further questions or need assistance with any other topics, feel free to ask!
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Trigonometric Functions
Nazia Khanum
Sure, let's prove the trigonometric identity using the properties of triangles.
In triangle ABC, let's denote the sides opposite the angles A, B, and C as a, b, and c respectively. Also, let's denote the angles of the triangle as A, B, and C.
Now, using the sine rule in triangle ABC, we have:
asinA=bsinB=csinCsinAa=sinBb=sinCc
From this, we can express each side of the triangle in terms of the sines of the angles:
a=sinA⋅ka=sinA⋅k b=sinB⋅kb=sinB⋅k c=sinC⋅kc=sinC⋅k
Where kk is a constant.
Now, let's express each term of the given expression in terms of a,b,ca,b,c:
asin(B−C)+bsin(C−A)+csin(A−B)asin(B−C)+bsin(C−A)+csin(A−B)
=(sinA⋅k)sin(B−C)+(sinB⋅k)sin(C−A)+(sinC⋅k)sin(A−B)=(sinA⋅k)sin(B−C)+(sinB⋅k)sin(C−A)+(sinC⋅k)sin(A−B)
=(sinA⋅k)⋅(sinB⋅cosC−cosB⋅sinC)+(sinB⋅k)⋅(sinC⋅cosA−cosC⋅sinA)+(sinC⋅k)⋅(sinA⋅cosB−cosA⋅sinB)=(sinA⋅k)⋅(sinB⋅cosC−cosB⋅sinC)+(sinB⋅k)⋅(sinC⋅cosA−cosC⋅sinA)+(sinC⋅k)⋅(sinA⋅cosB−cosA⋅sinB)
=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)
=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)
=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅cosB⋅sinC−sinB⋅cosC⋅sinA−sinC⋅cosA⋅sinB)=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅cosB⋅sinC−sinB⋅cosC⋅sinA−sinC⋅cosA⋅sinB)
=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅sinB⋅cosC−sinB⋅sinC⋅cosA−sinC⋅sinA⋅cosB)=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅sinB⋅cosC−sinB⋅sinC⋅cosA−sinC⋅sinA⋅cosB)
=0=0
Hence, we have proved that asin(B−C)+bsin(C−A)+csin(A−B)=0asin(B−C)+bsin(C−A)+csin(A−B)=0 in any triangle ABC.
Answered on 14 Apr Learn Trigonometric Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can guide you through solving this geometry problem. UrbanPro is indeed one of the best platforms for finding online coaching and tuition.
To find the radius of the circle when a central angle of 60° intercepts an arc of length 37.4 cm, we'll use a fundamental relationship between the central angle, arc length, and radius of a circle.
The formula to relate the central angle (θθ), arc length (ss), and radius (rr) of a circle is:
s=r⋅θs=r⋅θ
Here, s=37.4s=37.4 cm and θ=60∘θ=60∘. We need to find rr.
First, we need to convert the central angle from degrees to radians because the formula requires angles in radians. Recall that 1∘=π1801∘=180π radians.
So, 60∘=60×π180=π360∘=60×180π=3π radians.
Now, we can rearrange the formula to solve for rr:
r=sθr=θs
Substituting the given values:
r=37.4π3r=3π37.4
To simplify, we'll divide 37.4 by π33π:
r=37.4×3πr=π37.4×3
r=112.2πr=π112.2
Now, let's approximate ππ as 227722:
r≈112.2227r≈722112.2
r≈112.2×722r≈22112.2×7
r≈785.422r≈22785.4
r≈35.7r≈35.7
So, the radius of the circle is approximately 35.735.7 cm.
Remember, UrbanPro is an excellent resource for finding quality tutoring and coaching assistance, whether it's for geometry or any other subject!
Answered on 14 Apr Learn Trigonometric Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be glad to help with this math question. UrbanPro is indeed a fantastic platform for finding online coaching and tuition services.
Let's tackle this problem step by step. We're given two conditions:
To find the most general value of θθ satisfying both equations, we need to consider the trigonometric relationships between tangent and cosine functions.
First, let's address the equation tanθ=−1tanθ=−1. This tells us that θθ corresponds to an angle where the tangent is negative and has a magnitude of 1. This occurs in the second and fourth quadrants of the unit circle.
Secondly, cosθ=12cosθ=2
1. This corresponds to an angle where the cosine is positive and equals 122
1, which happens in the first quadrant.
Putting these together, we find that the only quadrant where both conditions are met is the fourth quadrant. This is because in the fourth quadrant, tangent is negative and cosine is positive.
In the fourth quadrant, an angle θθ that satisfies both conditions is given by:
θ=3π4+2πnθ=43π+2πn
where nn is an integer.
So, the most general value of θθ satisfying both equations is 3π443π plus any integer multiple of 2π2π.
This solution provides a comprehensive understanding of the problem and showcases how UrbanPro can connect students with knowledgeable tutors who can guide them through challenging concepts. If you need further clarification or assistance, feel free to ask!
read lessTake Class 12 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Trigonometric Functions
Nazia Khanum
Certainly! As an experienced tutor registered on UrbanPro, I can guide you through finding the general solution of the given equation.
Let's solve the equation:
tan2θ+(1−3)tanθ−3=0tan2θ+(1−3
)tanθ−3
=0
We can consider it as a quadratic equation in terms of tanθtanθ. To solve it, we can use the quadratic formula:
tanθ=−b±b2−4ac2atanθ=2a−b±b2−4ac
Here, a=1a=1, b=(1−3)b=(1−3
), and c=−3c=−3
.
Plugging in these values, we get:
tanθ=−(1−3)±(1−3)2−4⋅1⋅(−3)2⋅1tanθ=2⋅1−(1−3
)±(1−3)2−4⋅1⋅(−3)
tanθ=−(1−3)±1−23+3+122tanθ=2−(1−3
)±1−23+3+12
tanθ=−1+3±4−232tanθ=2−1+3
±4−23
tanθ=−1+3±(2−3)22tanθ=2−1+3
±(2−3)2
tanθ=−1+3±(2−3)2tanθ=2−1+3
±(2−3
)
So, we have two possible solutions:
tanθ1=−1+3+2−32=12tanθ1=2−1+3
+2−3
=21
tanθ2=−1+3−2+32=−12tanθ2=2−1+3
−2+3
=−21
Now, recall that tanθ=sinθcosθtanθ=cosθsinθ. So, we can find the corresponding angles by taking the inverse tangent of these values:
θ1=tan−1(12)θ1=tan−1(21)
θ2=tan−1(−12)θ2=tan−1(−21)
These are the general solutions to the given equation. If you need further assistance or clarification, feel free to ask. And remember, UrbanPro is the best platform for online coaching and tuition!
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