Volume of a cylinder with radius r and height h is \(\pi\)r²h

32 years and 40 years

(l)\(\rightarrow\)VERTICES
(2)\(\rightarrow\)HEXAGON
(3)\(\rightarrow\)EDGE
(4)\(\rightarrow\)QUADRILATERAL
(5)\(\rightarrow\)PENTAGON
(6)\(\rightarrow\)RHOMBUS
(7)\(\rightarrow\)POLYGON
(8)\(\rightarrow\)DIAGONAL

We have
A rational number between \(\frac { 1 }{ 2 } \)
= \(\left[ \frac { 1 }{ 2 } +(-2) \right] \div 2=\left[ \frac { 1-4 }{ 2 } \right] \div 2\)
= \(\left[ \frac { -3 }{ 2 } \right] \times \frac { 1 }{ 2 } =\frac { -3 }{ 4 } \)
A rational number between \(\frac { 1 }{ 2 } \) and \(\left( \frac { -3 }{ 4 } \right) \)
= \(\left[ \frac { 1 }{ 2 } +\left( \frac { -3 }{ 4 } \right) \right] \div 2\)
\(\left[ \frac { 2-3 }{ 4 } \right] \times \frac { 1 }{ 2 } =\frac { -1 }{ 4 } \times \frac { 1 }{ 2 } =\frac { -1 }{ 8 } \)
A rational number between \(\left( \frac { -3 }{ 4 } \right) \) and (-2)
= \(\left[ \left( \frac { -3 }{ 4 } \right) +(-2) \right] \div 2=\left[ \frac { (-3)+(-8) }{ 4 } \right] \times \frac { 1 }{ 2 } \)
= \(\frac { -11 }{ 4 } \times \frac { 1 }{ 2 } =\frac { -11 }{ 8 } \)
Thus, the three rational numbers
\(\left( \frac { -3 }{ 4 } \right) ,\left( \frac { -1 }{ 8 } \right) \) and \(\left( \frac { -11 }{ 8 } \right) \) are between \(\frac { 1 }{ 2 } \) and (-2)

a = 1

(i) (ab + bc) (ab - bc) + (bc + ca) (be - ca)+(ca+ab)(ca-ab)=0
LHS = (ab + bc)(ab - bc) + (bc + ca)(bc - ca)+ (ca + ab)(ca - ab)
= {(ab)² -(bc)²} + {(bc)² -(ca)²} + {(ca)² - (ab)²}
[by using identity, (A² - B²) = (A + B)(A - B)]
= a²b² -b²c² + b²c²-c²a² + c²a²-a²b²
= a²b² -a²b² + b²c² -b²c² + c²a² -c²a²
= 0= RHS
LHS = RHS
(ii) (a + b +c)(a² + b² + c² -ab-bc -ca)
= a³ + b³ + c³ - 3abc
LHS
= (a + b + c) \(\times\) (a² + b² + c²- ab - bc - ca)
= a³ + ab² + ac² -a²b - abc - ca²+ ba² + b³ + bc²- ab² - b²c - abc+ ca² + b²c + c³ - abc - bc² - ac²
= a³ + b³ + c³-abc - abc - abc + ab²- ab²+ ac² -ac² + a²b -a²b + ca² -ca²+ bc² - bc² + b²c - b²c
= a³ + b³ + c³- 3abc = RHS
LHS = RHS

4 pumps

(a) 5500
(b) 28000

Given,\(\ x+\frac { 1 }{ x } =5\)
\(\ \left( x+\frac { 1 }{ x } \right) ^{ 2 }=25\)
\(\left( x+\frac { 1 }{ x } \right) ^{ 2 }={ x }^{ 2 }+2\times x\times \frac { 1 }{ x } +\left( \frac { 1 }{ x } \right) ^{ 2 }\quad \)
(a+ b)² =a² + 2ab + b². Here, a = x and \(b=\frac { 1 }{ x } \)
\(={ x }^{ 2 }+2+\left( \frac { 1 }{ { x }^{ 2 } } \right) ={ x }^{ 2 }+\left( \frac { 1 }{ { x }^{ 2 } } \right) +2\)
\(\left( x+\frac { 1 }{ x } \right) ^{ 2 }=25\)
\(\quad { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2=25\)
\({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =25-2=23\)

Rs 330

Let ABCD be a given quadrilateral, which is a parallelogram. Join the diagonal BD of the parallelogram ABCD and it divides the parallelogram into two ΔABD and ΔBCD.

Then, area of parallelogram ABCD = Area of ABD + Area of ΔBCD.
 = \(\frac { 1 }{ 2 } \)\(\times\)AB\(\times\) h + CD\(\times\) h
=  \(\frac { 1 }{ 2 } \)\(\times\) b \(\times\)h + b \(\times\)h
=  \(\frac { bh }{ 2 } +\frac { bh }{ 2 } =\frac { bh+bh }{ 2 } =\frac { 2bh }{ 2 } \) = bh sq units
We know that,
Area of parallelogram = Base \(\times\) Height
 = b \(\times\)h = bh sq units
We also know that, a parallelogram can also be a trapezium.
∴ Area of trapezium ABCD =\(\frac { 1 }{ 2 } \)  \(\times\)(Sum of parallel lines) \(\times\)(Perpendicular distance between the parallel sides)
=\(\frac { 1 }{ 2 } \)\(\times\) (b + h) \(\times\)h
=\(\frac { 1 }{ 2 } \) \(\times\)2b\(\times\)h = bh sq units
Hence, we can say that the above relation agrees with formula that we know already.

Since 6 x 6 = 36
\(\therefore\) The unit digit of (12796) will be 6.

Let us connvert the given rational numbers having the same denominations
We have \(\frac { 5 }{ 8 } =\frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \) and \(\frac { -5 }{ 6 } \)
= \(\frac { -5\times 4 }{ 6\times 4 } =\frac { -20 }{ 24 } \)
Now, the rational numbers between \(\frac { -20 }{ 24 } \) and \(\frac { 15 }{ 24 } \) are \(\frac { -19 }{ 24 } ,\frac { -18 }{ 24 } ,\frac { -17 }{ 24 } ,....\frac { -1 }{ 24 } ,\frac { 0 }{ 24 } ,\frac { 1 }{ 24 } ,\frac { 2 }{ 24 } ,\frac { 3 }{ 24 } ,...,\frac { 14 }{ 24 } \)
We can take any seven of these

\((5x^{2}-6x)\div 3x\)=\(\frac{5x^{2}-6x}{3x}=\frac{5x^{2}}{3x}-\frac{6x}{3x}\)
\(=\frac{5}{3}x-2=\frac{5x-6}{3}=\frac{1}{3}(5x-6)\)

2.4 m

255xy³z²

Length of longer side = 34 m
Length of shorter side = 13 m

Let the length of the rectangle = 1 cm
\(\therefore\) Since width of the rectangle (b) = \(3{3\over 4}\) and perimeter = 17 cm
We know that perimeter of a rectangle = 2(l + b) cm
Then, \(2(l+3{3\over 4})=17\) or \(2(l+{15\over 4})=17\)
or \(2l+(2\times {15\over 4})=17\) or \(2l+{15\over 2}=17\)
or \(2l=17-{15\over 2}\) (Transposing \({15\over 2}\) to RHS)
or \(2l={35-15\over 2}={19\over 2}\)
Dividing both side by 2, we have
\(i={19\over 2}\times {1\over 2}={19\over 4}=4{3\over 4}\) cm
\(\therefore\) Length of the rectangle = \(4{3\over 4}\)cm.

180 cm²

Here, ab = 21 and a + b = -10
Possible values of a and b are 7, 3 or -7, -3.
But  7 +3 = 10 \(\neq \) - 10     [not possible]
\(\therefore\) a= -7, b = -3
Now q² - 10q + 21 = q² +(-7-3)q + 21
= q² - 7q - 3q + 21
=q (q  -7)-3(9 - 7) = (q-7)(q -3)
Hence, required factors of given expression are(q - 7) and (q -3).

Principal for the third year, (P) = Amount at the end of twoyears = Rs 8820, rate (R) = 5%, time (T) = 1yr
∴ Simple interest = \(\frac{P\times R\times T}{100}=\frac{8820\times5\times1}{100}\)
=\(\frac{882}{2}\) = Rs 441
Hence, interest for the third year is Rs 441.

1000 ÷ 10 = 100 = 10²

Area = \(\frac { 6\times 3 }{ 2 } \)= 9 cm²

\({30\over200}={12\over?}⇒?=80\)

Coefficient = 7.

x + y = 8
y + 5 = 10 \(\Rightarrow\) y = 5
\(\therefore\) x + 5 =.8 \(\Rightarrow\) x = 3.

Total number of students = 7 + 8 + 4 + 9 + 2 = 30.

0\(\times\)0\(\times\)0=0.