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To construct the quadrilateral PQRS with the given specifications, follow these steps:

1. Start by drawing a line segment PQPQ of length 5.4 cm. This will be one side of the quadrilateral.

2. At point PP, construct an angle of 60 degrees. To do this, use a protractor to measure and mark an angle of 60 degrees from line segment PQPQ.

3. At point QQ, construct an angle of 105 degrees. To do this, use a protractor to measure and mark an angle of 105 degrees from line segment PQPQ.

4. From point QQ, draw a ray extending from the angle bisector of angle QQ. This will be side QRQR.

5. At point RR, construct an angle of 75 degrees. To do this, use a protractor to measure and mark an angle of 75 degrees from line segment QRQR.

6. From point RR, draw a ray extending from the angle bisector of angle RR. This will be side RSRS.

7. At point SS, construct an angle of 120 degrees. To do this, use a protractor to measure and mark an angle of 120 degrees from line segment RSRS.

8. Finally, draw a line segment connecting point SS to point PP.

By following these steps, you will have constructed the quadrilateral PQRS with the given specifications.

A triangular pyramid, also known as a tetrahedron, is a type of pyramid that has a triangular base and three triangular faces that meet at a single vertex.

A pyramid with a square base is called a square pyramid. It has a square base and four triangular faces that meet at a single vertex, which is directly above the center of the square base.

Sure, let's draw the net of the dice, keeping in mind that the numbers on opposite faces must total 7.

Here's how you can draw the net:

1. Start with a square. Label the vertices of the square as A, B, C, and D.

2. Draw two lines perpendicular to each side of the square, dividing each side into two equal segments. Label the points of intersection with the sides as E, F, G, and H.

3. Join the vertices E, F, G, and H to the opposite vertices of the square, creating four triangles.

4. Label the vertices of these triangles with the numbers from 1 to 6 such that opposite faces sum to 7:

   • Opposite faces: A and C, B and D, E and G, F and H.

   A: 1 C: 6

   B: 2 D: 5

   E: 3 G: 4

   F: 3 H: 4

Now, you have the net of the dice, where the numbers on opposite faces total 7.

To show that a palindrome of 4 digits is divisible by 11, let's consider a generic 4-digit palindrome:

Let the palindrome be represented as abbaabba, where aa and bb are digits.

The value of this palindrome is 1000a+100b+10b+a1000a+100b+10b+a.

Simplifying, we get 1001a+110b1001a+110b.

For the palindrome to be divisible by 11, the difference between the sum of the digits in the even-numbered places and the sum of the digits in the odd-numbered places should be divisible by 11.

The difference between the sum of the digits in the even-numbered places (a+ba+b) and the sum of the digits in the odd-numbered places (a+ba+b) is 0.

Since the difference is 0, the palindrome is divisible by 11.

Thus, any 4-digit palindrome is divisible by 11.

As an experienced tutor registered on UrbanPro.com, I specialize in providing high-quality online coaching for Class 7 Tuition. One of the topics frequently covered in this grade is algebraic expressions and factorization. In this response, I will address the specific factorization question: "Factorise: 16x⁴ – y⁴."

Solution:

• Step 1: Identify the Perfect Square Form:

  • Recognize that the given expression is in the form of a difference of squares.

• Step 2: Apply the Difference of Squares Formula:

  • Utilize the formula a2−b2=(a+b)(a−b)a2−b2=(a+b)(a−b) where a=4x2a=4x2 and b=y2b=y2.

• Step 3: Substitute and Simplify:

  • Substitute the values of aa and bb into the formula.
  • Factorize 16x4−y416x4−y4 as (4x2+y2)(4x2−y2)(4x2+y2)(4x2−y2).

• Step 4: Further Factorization if Possible:

  • Notice that (4x2−y2)(4x2−y2) is another difference of squares.
  • Apply the difference of squares formula again: (4x2−y2)=(2x+y)(2x−y)(4x2−y2)=(2x+y)(2x−y).

Final Factorization: The complete factorization of 16x4−y416x4−y4 is (4x2+y2)(2x+y)(2x−y)(4x2+y2)(2x+y)(2x−y).

Conclusion: For effective Class 7 Tuition and clear explanations of concepts like factorization, consider enrolling in my online coaching sessions on UrbanPro.com. My goal is to provide comprehensive support to students, helping them grasp mathematical concepts with ease.

To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that:

(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq

In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which is 6) and pqpq to be equal to the constant term (which is 8).

Let's find pp and qq:

p+q=6p+q=6 pq=8pq=8

The pairs of numbers that satisfy these conditions are p=2p=2 and q=4q=4.

Therefore, the factorization is:

x2+6x+8=(x+2)(x+4)x2+6x+8=(x+2)(x+4)

The given expression is a difference of squares, which can be factored as follows:

49y2−1=(7y)2−1249y2−1=(7y)2−12

Now, you can use the difference of squares formula, which states that a2−b2=(a+b)(a−b)a2−b2=(a+b)(a−b). In this case, let a=7ya=7y and b=1b=1:

(7y+1)(7y−1)(7y+1)(7y−1)

So, the factorization of 49y2−149y2−1 is (7y+1)(7y−1)(7y+1)(7y−1).

To divide the expression 10(x3y2z2+x2y3z2+x2y2z3)10(x3y2z2+x2y3z2+x2y2z3) by 5x2y2z25x2y2z2, you can simplify by dividing each term in the numerator by the denominator:

Now, simplify each term:

1. 10x3y2z25x2y2z2=2x3−2y2−2z2−2=25x2y2z210x3y2z2=2x3−2y2−2z2−2=2
2. 10x2y3z25x2y2z2=2x2−2y3−2z2−2=25x2y2z210x2y3z2=2x2−2y3−2z2−2=2
3. 10x2y2z35x2y2z2=2x2−2y2−2z3−2=2z5x2y2z210x2y2z3=2x2−2y2−2z3−2=2z

Combine the simplified terms:

2+2+2z2+2+2z

So, the result of the division is 4+2z4+2z.