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  • 1 answers

Shivanya R 4 years, 2 months ago

Thumaara naam kyaa he betas?
  • 4 answers

Shehrebanu .... 4 years, 2 months ago

Varthana....... two different variables can never be added or subtracted
X²- 2y

Shivanya R 4 years, 2 months ago

2x_2y = 0xy

Shivanya R 4 years, 2 months ago

Very eas6
  • 3 answers

Meenu Gupta 4 years, 2 months ago

43/90

Shine Rai 4 years, 2 months ago

How, 43/90

Gaurav Seth 4 years, 2 months ago

Let 'x' be 0.47(bar on 7).

Since the number of repeating decimal is 1,we should multiply x and it's value with 10.

10*x=10*0.47(bar on 7).
10x=4.77(bar on 7).

Now, we should subtract x and it's value from 10x and it's value.

10x-x=4.77(bar on 7)-0.47(bar on 7).
9x=4.3(bar gets cancelled).
x=4.3/9.

This still needs to be multiplied with 10 in order to make p also an integer.

43/90 is the p/q form of 0.47(bar on 7) where p and q are integers and q is mot equal to 0.

  • 4 answers

Shehrebanu .... 4 years, 2 months ago

104 * 96 =(100+4) - (100-4) =100^2 - 4^2 =10000 - 16 =9984
9984

Shine Rai 4 years, 2 months ago

U can do this in more short

Yogita Ingle 4 years, 2 months ago

104×96

= (100 + 4) 96

= 96 (100) + 96(4)

= 9600 + 384
= 9984

  • 1 answers

Dharmesh Jagrat 4 years, 2 months ago

Rational number equavalent to 5/7 are 5/7 × 2/2 = 10/14 5/7 × 3/3 = 15/21 5/7 × 4/4 = 20/28 5/7 × 5/5 = 25/37
  • 3 answers

Shine Rai 4 years, 2 months ago

A term or a constant having whole number in power.

Rishikesh Patidar 4 years, 2 months ago

A term that has 4 or more than 4 terms is called polynomial

Ash Greninja 4 years, 2 months ago

A polynomial is an expression having an more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
  • 1 answers

Yogita Ingle 4 years, 2 months ago

In order to prove that the sum of angles of a triangle is 180, you must know the theorems of angles of a triangle.

We know that,  alternate interior angles are of equal magnitude.

This'll help us get the answer. Let us assume a triangle ABC. Now we have to substitute the angles.

∠PAB + ∠BAC + ∠CAQ = 180°. where PQ line parallel to side BC touching vertex A.

∠PAB = ∠ABC & ∠CAQ = ∠ACB BECAUSE alternate interior angles are congruent..

Hence we get, ∠ABC + ∠BAC + ∠ACB = 180°   [Proved]

  • 1 answers

Ayush Tiwari 4 years, 2 months ago

It's Probability
  • 4 answers

Shehrebanu .... 4 years, 2 months ago

A method other than equaling denominators is to add 1 to how much rational no. you want to find..........and then convert the denominator according to the sum we got by adding 1...... ..and then write the rational no. between the two no. you got.......and that you got the answer ☺

Arjun Prabhu V.K 4 years, 2 months ago

Thanks

Sreya Dutta 4 years, 2 months ago

First do the lcm of denominators.. Then convert the fraction to equivalent fractions containing the changed denominators.. After having same denominators.. Write fractions between those changed fractions. Ok?

Namish Jindal 4 years, 2 months ago

Pata nahi
  • 1 answers

Ash Greninja 4 years, 2 months ago

For Fig. 4. 6 (i) y = x (ii) x + y = 0 (iii) y = 2x (iv) 2 + 3y = 7x Solution: The points given in the figure 4.6 are (0,0), (-1,1), (1,-1) Substituting the values for x and y from these points in the equations, we get, (i) y = x (0,0) ⟹ 0 = 0 (-1, 1) ⟹ -1≠1 --------------------------- equation not satisfied (1, -1) ⟹ 1≠ -1 --------------------------- equation not satisfied (ii) x + y = 0 (0,0) ⟹ 0 + 0 = 0 (-1, 1) ⟹ -1+1= 0 (1, -1) ⟹ 1+ (-1)=0 (iii) y = 2x (0,0) ⟹ 0 = 2× 0 0 = 0 (-1, 1) ⟹ 1= 2×(-1) 1 ≠ -2 --------------------------- equation not satisfied (1, -1) ⟹ -1=2×1 -1≠2 --------------------------- equation not satisfied (iv) 2 + 3y = 7x (0,0) ⟹ 2+(3×0) = 7× 0 2 ≠ 0 --------------------------- equation not satisfied (-1, 1) ⟹ 2+(3×1) = 7× -1 5 ≠ -7 --------------------------- equation not satisfied (1, -1) ⟹ 2+(3×-1)= 7× 1 -1≠ 7 --------------------------- equation not satisfied Since, only equation x + y = 0 satisfies all the points, the equation whose graphs are given in Fig. 4.6 is x + y = 0 For Fig. 4. 7 (i) y = x + 2 (ii) y = x – 2 (iii) y = –x + 2 (iv) x + 2y = 6 Solution: The points given in the figure 4.7 are (0,2), (2,0), (-1,3) Substituting the values for x and y from these points in the equations, we get, (i) y = x + 2 (0,2) ⟹ 2 = 0+2 2 = 2 (2, 0) ⟹ 0= 2+2 0≠4 --------------------------- equation not satisfied (-1, 3) ⟹ 3= -1+2 3≠1 --------------------------- equation not satisfied (ii) y = x – 2 (0,2) ⟹ 2 = 0 –2 2 ≠ – 2 --------------------------- equation not satisfied (2, 0) ⟹ 0= 2–2 0= 0 (-1, 3) ⟹ 3= –1–2 3≠ –3 --------------------------- equation not satisfied (iii) y = –x + 2 (0,2) ⟹ 2 = -0+2 2 = 2 (2, 0) ⟹ 0 = -2+2 0=0 (-1, 3) ⟹ 3= -(-1)+2 3=3 (iv) x + 2y = 6 (0,2) ⟹ 0+(2× 2) = 6 4 ≠ 6 --------------------------- equation not satisfied (2, 0) ⟹ 2+(2× 0) = 6 2 ≠ 64 --------------------------- equation not satisfied (-1, 3) ⟹ -1+(2× 3) = 6 5≠6 --------------------------- equation not satisfied Since, only equation y = –x + 2 satisfies all the points, the equation whose graphs are given in Fig. 4.7 is y = –x + 2.
  • 1 answers

Shehrebanu .... 4 years, 2 months ago

For Fig. 4. 6 (i) y = x (ii) x + y = 0 (iii) y = 2x (iv) 2 + 3y = 7x Solution: The points given in the figure 4.6 are (0,0), (-1,1), (1,-1) Substituting the values for x and y from these points in the equations, we get, (i) y = x (0,0) ⟹ 0 = 0 (-1, 1) ⟹ -1≠1 --------------------------- equation not satisfied (1, -1) ⟹ 1≠ -1 --------------------------- equation not satisfied (ii) x + y = 0 (0,0) ⟹ 0 + 0 = 0 (-1, 1) ⟹ -1+1= 0 (1, -1) ⟹ 1+ (-1)=0 (iii) y = 2x (0,0) ⟹ 0 = 2× 0 0 = 0 (-1, 1) ⟹ 1= 2×(-1) 1 ≠ -2 --------------------------- equation not satisfied (1, -1) ⟹ -1=2×1 -1≠2 --------------------------- equation not satisfied (iv) 2 + 3y = 7x (0,0) ⟹ 2+(3×0) = 7× 0 2 ≠ 0 --------------------------- equation not satisfied (-1, 1) ⟹ 2+(3×1) = 7× -1 5 ≠ -7 --------------------------- equation not satisfied (1, -1) ⟹ 2+(3×-1)= 7× 1 -1≠ 7 --------------------------- equation not satisfied Since, only equation x + y = 0 satisfies all the points, the equation whose graphs are given in Fig. 4.6 is x + y = 0 For Fig. 4. 7 (i) y = x + 2 (ii) y = x – 2 (iii) y = –x + 2 (iv) x + 2y = 6 Solution: The points given in the figure 4.7 are (0,2), (2,0), (-1,3) Substituting the values for x and y from these points in the equations, we get, (i) y = x + 2 (0,2) ⟹ 2 = 0+2 2 = 2 (2, 0) ⟹ 0= 2+2 0≠4 --------------------------- equation not satisfied (-1, 3) ⟹ 3= -1+2 3≠1 --------------------------- equation not satisfied (ii) y = x – 2 (0,2) ⟹ 2 = 0 –2 2 ≠ – 2 --------------------------- equation not satisfied (2, 0) ⟹ 0= 2–2 0= 0 (-1, 3) ⟹ 3= –1–2 3≠ –3 --------------------------- equation not satisfied (iii) y = –x + 2 (0,2) ⟹ 2 = -0+2 2 = 2 (2, 0) ⟹ 0 = -2+2 0=0 (-1, 3) ⟹ 3= -(-1)+2 3=3 (iv) x + 2y = 6 (0,2) ⟹ 0+(2× 2) = 6 4 ≠ 6 --------------------------- equation not satisfied (2, 0) ⟹ 2+(2× 0) = 6 2 ≠ 64 --------------------------- equation not satisfied (-1, 3) ⟹ -1+(2× 3) = 6 5≠6 --------------------------- equation not satisfied Since, only equation y = –x + 2 satisfies all the points, the equation whose graphs are given in Fig. 4.7 is y = –x + 2.
  • 1 answers

Alok Gupta 4 years, 2 months ago

The root of a number x is another number, which when multiplied by itself a given number of times, equals x. For example the second root of 9 is 3, because 3x3 = 9. The second root is usually called the square root. The third root is usually called the cube root.
  • 2 answers

Gaurav Seth 4 years, 2 months ago

A n s w e r:

x = 2 , y = 5

S t e p-b y-s t e p e x  p l a n a t i o n:

x = 2

2y-3 = 7

2y = 10

y = 5

Mufeedha S 4 years, 2 months ago

with explaination
  • 1 answers

Gaurav Seth 4 years, 2 months ago

• The axiom is a statement which is self evident.

But,a theorem is a statement which is not self evident.

 

• An axiom cannot be proven by any kind of mathematical representation.

But, a theorem can be proved by mathematical representation.

 

• A theorem can be proved or derived from the axioms.

But,axioms cannot be proven or derived by the theorems.

  • 1 answers

Silviya Shrivastav 4 years, 2 months ago

I hope you will be able to make the diagram: Let the Angles be =x In∆ABC x+x+50°=180°(Angle Sum Property) 2x+50 =180° 2x=180°-50° 2x=130 x=130°/2 x=65° PQ||BC Angle C+AngleQ=180°(co-interior angle) 65°+Angle PQC=180° AnglePQC=180°-65° PQC=115° HOPE IT HELPS!!
  • 5 answers

Mahira Srivastava 4 years, 2 months ago

Its 22/4 or 3.14 Hope its helps you !

Agent Nobody 4 years, 2 months ago

22/7

Garima Chaudhary 4 years, 2 months ago

3.14 or 22/7

Cm| Black Bolt 4 years, 2 months ago

22/7 (Fraction Form) 3.14 (Decimal Form) Hope this helps you

Meenu Gupta 4 years, 2 months ago

The value can be 22/7 or 3.14
  • 3 answers

Mahira Srivastava 4 years, 2 months ago

Its 115

Meenu Gupta 4 years, 2 months ago

Answer is 115

Cm| Black Bolt 4 years, 2 months ago

The answer is 115
  • 2 answers

Yogita Ingle 4 years, 2 months ago

2x - 9x + 4
= 2x - 8x - x + 4

= 2x ( x - 4) - 1 ( x - 4)

= (x - 4) (2x - 1)

Femely Shaji 4 years, 2 months ago

Step by step plzzz
  • 1 answers

Yogita Ingle 4 years, 2 months ago

Given,

p(x) = 2x²-3x+7a

2 is the zero of the polynomial

So, substitution

p(2) = 0

p(2) = 2(2)2 - 3x + 7a

0 = 2(4)-3(2)+7a

0 = 8-6+7a

0 = 2+7a

-2 = 7a

-2/7 = a

-2/7 = a

The value of a is -2/7

  • 2 answers

Roshan Chand 4 years, 2 months ago

Mean is the sum of terms or numbers / no of terms 23 + 29 / 2 = 52 / 2 = 26 So the correct answer is d)26

Roshan Chand 4 years, 2 months ago

Mean is sum of terms or numbers/ no of terms 20 + 30 / 2 = 50 / 2 = 25 So the answer is a)25
  • 3 answers

Meenu Gupta 4 years, 2 months ago

50

Mahira Srivastava 4 years, 2 months ago

50

Yogita Ingle 4 years, 2 months ago

Given class interval (40-60)

Lower limit = 40

Upper limit = 60

Class mark = (40 + 60)/2
= 100/2

= 50

Therefore, Class mark of the class interval 40-60 is 50.

  • 1 answers

Gaurav Seth 4 years, 2 months ago

Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.

Join the diagonal BD which divides quadrilateral ABCD in two triangles i.e ∆BCD & ∆ABD.

In ΔBCD,
By applying Pythagoras Theorem

BD²=BC² +CD²

BD²= 12²+ 5²= 144+25

BD²= 169
BD = √169= 13m

∆BCD is a right angled triangle.

Area of ΔBCD = 1/2 ×base× height

=1/2× 5 × 12= 30 m²

For ∆ABD,
Let a= 9m, b= 8m, c=13m

Now,
Semi perimeter of ΔABD,(s) = (a+b+c) /2

s=(8 + 9 + 13)/2 m
= 30/2 m = 15 m

s = 15m

Using heron’s formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 – 9) (15 – 9) (15 – 13)

= √15 × 6 × 7× 2
=√5×3×3×2×7×2
=3×2√35
= 6√35= 6× 5.92

[ √6= 5.92..]
= 35.52m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD
= 30+ 35.5= 65.5 m²

Hence, area of the park is 65.5m²

  • 0 answers
  • 3 answers

Yogita Ingle 4 years, 2 months ago

5x2 + 16x + 3 = 0
5x2 + 15x + x + 3 = 0

5x ( x + 3) + 1 (x + 3) =0

(x + 3) (5x + 1) =0

x + 3   = 0 or 5x + 1 =0

x = -3  or 5x = -1

x = -3 or x = -1/5

Femely Shaji 4 years, 2 months ago

Step b step plzzz

Rohan Solanki 4 years, 2 months ago

=0
  • 0 answers
  • 2 answers

Gaurav Seth 4 years, 2 months ago

<pre> (i) p(x) = 5x² – 3x + 7 at x = 1. p(1) = 5 x (1)² - 3 x 1 + 7 = 5 - 3 + 7 = 12 - 3 = 9 </pre>

Shruti Joshi 4 years, 2 months ago

9
  • 2 answers

Shruti Joshi 4 years, 2 months ago

0.230769230769....

Yogita Ingle 4 years, 2 months ago

Decimal form of 3/13 is 0.2307692. Decimal expansion is the Terminating and repeating which is about 9th numbers is repeating it.

Converting the fraction to the decimal which is easily dividing numerator by denominator is the correct answer that is 0.2307692.

Get 3/13 converted to decimal with dividing 3 by 13.3 /13 as a decimal are then converted into the above aspects.

  • 1 answers

Cm| Black Bolt 4 years, 2 months ago

0 divided by any number is 0. Solution for your question
  • 0 answers

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