CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Side or the tri. = 240÷3=80. now, s=a+b+c/2 = 240/2 =80 Now, put herons formula Then u will get answer Like this)) 2x2x2x2x2x2x5x5xroot 3 =1600root 3 Thats the answer..... :)

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Can u give me blue print of maths?

Posted by Mansi Asati 7 years, 9 months ago

CBSE > Class 10 > Mathematics

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Retika Malhotra 7 years, 9 months ago

Available in this app

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Aman Singh 7 years, 9 months ago

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How to do completing the square method ???

Posted by Jeno Pearlin 7 years, 9 months ago

CBSE > Class 10 > Mathematics

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Rajnish Yadav 7 years, 9 months ago

It's very simple fella follow these steps: Just suppose your equation is 2xsquare - 7x+3=0 Here a is 2 B is 7 C is 3 Now use formula D=(b)square -4ac If D>O and D=O then they exist Now here D is 25 so it exists.. Now divide both side of your equation by 2 and multiply the constant of x by 1/4 and add and subtract it on LHS of equation and also square them then at last find the values of x..

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Prove that Tanx/1-cotx + cotx/1-tanx = 1+tanx+cotA

Posted by Himanshu Pal 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Himanshu Pal 7 years, 9 months ago

Give me answer for this question

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what is origin ?

Posted by Golu Kumar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

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Abhinav Jha 7 years, 9 months ago

It has coordinates(0,0)

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Aman Singh 7 years, 9 months ago

The origin is the zero point on a number line (= drawing that represents all the numbers that exist), or the point where the horizontal and vertical lines on a graph (= drawing) cross.

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How to prove uderroot 3.5 is irrational number

Posted by Manish Prajapati 7 years, 9 months ago

CBSE > Class 10 > Mathematics

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An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameter of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.

Posted by Gaurav Pal 6 years, 6 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 6 months ago

Area of metallic sheet used = curved surface area of Frustum + curved surface area of Cylinder + area of circular Base.
Diameter of the bigger circular end = 45 cm
Radius, r₁ = {tex}\frac{45}2{/tex}= 22.5 cm {tex}{/tex}
Diameter of the smaller circular end = 25 cm
Radius, r₂ = {tex}\frac{{25}}{2} = 12.5\ cm{/tex}
Height of the frustum, h = Total height of the bucket (H) - Height of the circular base (h₁)
{tex}={/tex} 40 - 6 = 34 cm
Slant Height of frustrum (l) {tex}=\sqrt { h ^ { 2 } + \left( r_ 1 - r_ 2 \right) ^ { 2 } }{/tex}
{tex}\Rightarrow l= \sqrt { 34 ^ { 2 } + ( 22.5 - 12.5 ) ^ { 2 } }{/tex}
{tex}\Rightarrow l=\sqrt { 1156 + ( 10 ) ^ { 2 }}{/tex}
{tex}\Rightarrow l=\sqrt { 1156 + 100}{/tex}
{tex}\Rightarrow l=\sqrt { 1256 }{/tex}
{tex}\Rightarrow{/tex} Slant Height, l = 35.44 cm
Curved surface area of Frustum = {tex}\pi ( r_1 + r _2 )l{/tex}
= {tex}\frac{{22}}{7} \times (22.5 + 12.5) \times 35.44{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{{22}}{7} \times 35 \times 35.44{/tex}
= 3898.4 cm²
Area of Circular Base with radius {tex}\frac{{25}}{2} = 12.5{/tex} is given by,
Area of circular base = {tex}\pi r_2^2{/tex}
{tex}{/tex}{tex}=\frac{{22}}{7} \times 12.5 \times 12.5{/tex}
= 491.07 cm²
Curved surface area of Cylinder = {tex}2\pi r_2h_1{/tex}
{tex}={/tex} {tex}2 \times \frac{{22}}{7} \times 12.5 \times 6{/tex}
{tex}={/tex} 471.428 cm²
Area of metallic sheet used = 3898.4 cm² + 491.07 cm² + 471.428 cm²
= 4860.898 cm²
Volume of water in bucket = Volume of Frustrum
= {tex}\frac13\pi h(r_1^2\;+\;r_2^2\;+\;r_1r_2){/tex}
= {tex}\frac13\times\frac{22}7\times34\;(22.5^2\;+\;12.5^2\;+\;22.5\times12.5){/tex} = 33615.48 cm³
Now, 1 litre = 1000 cm
Thus, Volume of water in bucket in litres = 33.62 litres

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Find the roots of the equation x^2-3x-m (m+3)=0, where m is a constant.

Posted by . . 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

{tex}x^2 - 3x -m(m + 3) = 0 {/tex}
{tex}\Rightarrow{/tex}{tex} x^2 + mx - (m + 3)x - m(m + 3) = 0{/tex}
{tex}\Rightarrow{/tex}{tex}x(x+m)-(m+3)(x+m){/tex}
{tex}\Rightarrow{/tex} (x+m) (x - m - 3) = 0
x = -m and m + 3

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Show that the point (1,7) , (4,2) , (-1,-1) and (-4,4) are the vertices of a square

Posted by Mental Roy 7 years, 9 months ago

CBSE > Class 10 > Mathematics