Question 511 Mark
The standard form for 203000 is $2.03 × 10^5$.
AnswerTrue.
Solution:
For standard form, $203000 = 203 × 10 × 10 × 10 = 203 × 10^3$
$= 2.03 × 10^2× 10^3$
$= 2.03 × 10^5$
View full question & answer→Question 521 Mark
What will the following machine do to a $2\ cm$ long piece of chalk?
AnswerThe machine produce $x1^{100}= 1$
So, if we insert $2\ cm$ long piece of chalk in that machine, the piece of chalk remains same.
View full question & answer→Question 531 Mark
$\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}=$ _________.
Answer$\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}=\Big(\frac{2}{13}\Big)^{-36}$
Solution:
Using laws of exponents, $a^m \div a^n=(a)^{m-n}$ and $a^m \times a^n=(a)^{m+n}$[$\therefore$ a is non-zero integer]
$\therefore$ $\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}$
$=\bigg[\Big(\frac{2}{13}\Big)^{-6-3}\bigg]^3\times\Big(\frac{2}{13}\Big)^{-9}$
$=\Big(\frac{2}{13}\Big)^{-27}\times\Big(\frac{2}{13}\Big)^{-9}$
$=\Big(\frac{2}{13}\Big)^{-27-9}=\Big(\frac{2}{13}\Big)^{-36}$
Hence, $\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}=\Big(\frac{2}{13}\Big)^{-36}$
View full question & answer→Question 541 Mark
The expontential form for $(-2)^4\times\Big(\frac{5}{2}\Big)^4$ is $5^4$.
AnswerUsing law of exponents, $a^m \div a^n=(a)^{m-n}$ and $\Big(\frac{\text{a}}{\text{b}}\Big)^\text{m}=\frac{\text{a}^\text{m}}{\text{b}^\text{m}}$
[$\because a$ and $b$ are non-integers]
$\therefore$ $(-2)^4\times\Big(\frac{5}{2}\Big)^4=(2)^4\times\frac{(5)^4}{(2)^4}$
$=(2)^{4-4}\times5^4$
$=2^0\times5^4=5^4$ [$\because$ $(-a^m) = (a^m)$, if $m$ is an even number]
View full question & answer→Question 551 Mark
By multiplying $\Big(\frac{5}{3}\Big)^4$ by ________ we get $5^4$.
AnswerBy multiplying $\Big(\frac{5}{3}\Big)^4$ by ________ we get $5^4$.
Solution:
Let $x$ be multiplied with $\Big(\frac{5}{3}\Big)^4$ to get $5^4$.
So,
$\Big(\frac{5}{3}\Big)^4\times\text{x}=5^4$
$\therefore$ $\text{x}=\frac{5^4}{\Big(\frac{5}{3}\Big)^4}$
$=5^4\times3^4\times5^{-4}=(5)^{4-4}\times3^4$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}},\text{a}^\text{m}\times\text{a}^{\text{n}}=(a)^{\text{m}+\text{n}}\Big]$
$=5^o\times3^4=1\times81=81$ [$\because$ $a^0= 1$]
View full question & answer→Question 561 Mark
$\Big(-\frac{8}{2}\Big)^0=0$
Answer$LHS =\Big(-\frac{8}{2}\Big)^0$
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore$ $\Big(-\frac{8}{2}\Big)=1$
$LHS ≠ RHS$
View full question & answer→Question 571 Mark
The value for $(-7)^6 \div 7^6$ is _________.
Answer Using law of exponents, $a^m \div a^n=(a)^{m-n}[\because$ a is non-zero integer$]$
$\therefore(-7)^6 \div 7^6=(-7)^6 \div 7^6\left[\left(-a^m\right)=\left(a^m\right)\right.$, if $m$ is an even number$]$
$=(7)^{6-6}=(7)^0=1\left[\because a^0=1\right]$
Hence,
$(-7)^6 \div 7^6=1$
View full question & answer→Question 581 Mark
The expression for $4^{-3}$ as a power with the base $2$ is $2^6$.
AnswerFalse.Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ $\therefore$ $4^{-3}=\frac{1}{4^3}$ $=\frac{1}{(2^2)^3}=\frac{1}{(2)^6}$ $[\because2\times2=4,(\text{a}^\text{m})^\text{n}=(\text{a})^\text{mn}]$
View full question & answer→Question 591 Mark
$\Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}=\Big(\frac{2}{3}\Big)^{10}$
AnswerFalse.Solution:
$LHS \Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}$
Using law of exponents, $a^m \times a^n=(a)^{m+n}[\because$ a is non-zero integer$]$
$\therefore$ $\Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}=\Big(\frac{2}{3}\Big)^{-2-5}$
$=\Big(\frac{2}{3}\Big)^{-7}$
$LHS ≠ RHS$
View full question & answer→Question 601 Mark
Express $3^{-5}\times 3^{-4}$ as a power of $3$ with positive exponent.
AnswerUsing laws of exponents, $a^m \times a^n=(a)^{m+n}$ and $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}[\because$ a is not-zero integer$]$
$\therefore$ $3^{-5}\times3^{-4}=(3)^{-5-4}$
$=(3)^{-9}=\frac{1}{3^9}$
View full question & answer→Question 611 Mark
Supply the missing information for diagram.
AnswerIf $x\ cm$ long piece is inserted in $(x4)$ and $(x3)$ hooked machine, repeated machine, then it will produce $36\ cm$ long piece. So, $x \times 4 \times 3 = 36$
$ \Rightarrow x = 3\ cm$
View full question & answer→Question 621 Mark
Find three machines that can be replaced with hook-ups of $(x^5)$ machines.
AnswerSince, $5^2= 25, 5^3= 125, 5^4= 625$
Hence, $(x5^2), (x5^3)$ and $(x5^4)$ machine can replace $(x^5)$ hook-up machine.
View full question & answer→Question 631 Mark
By solving $(6^0- 7^0) × (6^0+ 7^0)$ we get ________.
AnswerUsing law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore$ $(6^0- 7^0) × (6^0+ 7^0) = (1 - 1) × (1 + 1)$
$= 0 × 2 = 0$
Hence,
$(6^0- 7^0) × (6^0+ 7^0) = 0$
View full question & answer→Question 641 Mark
If $a = -1, b = 2, $ then find the value of the following: $a^b÷ b^a$
AnswerGiven, $a^b+ b^a$
If $a = -1$ and $b = 2,$ then $(-1)^2 \div(-2)^{-1}=1+\frac{1}{2^1}={1}\times{2}=2$
$\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→Question 651 Mark
Use the properties of exponents to verify that each statement is true. $\frac{1}{4}(2^{\text{n}})=2^{\text{n}-2}$
Answer$\frac{1}{4}(2^{\text{n}})=2^{\text{n}-2}$ $\text{RHS} = 2^{\text{n} - 2} = 2^\text{n} + 2^2 $ $[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=(\text{a})^{\text{m}-\text{n}}]$ $=\frac{2^{\text{n}}}{4}=\text{LHS}$
View full question & answer→Question 661 Mark
Supply the missing information for diagram.
AnswerIf $5\ cm$ long piece is inserted in single machine, then it produce same $5\ cm$ long piece. So, it is $(x1)$ repeated machine. $\because \ ?=1$
View full question & answer→Question 671 Mark
The standard form of $\Big(\frac{1}{100000000}\Big)$ is ______.
AnswerThe standard form of $\Big(\frac{1}{100000000}\Big)$ is $1.0 \times 10^{-8}$
Solution:
For standard form, $\Big(\frac{1}{100000000}\Big)=\frac{1}{1\times10^8}=\frac{1}{10^8}$
$=1 \times 10^{-8}=1.0 \times 10^{-8}$
Hence, Standard form of $\Big(\frac{1}{100000000}\Big)$ is $1.0 \times 10^{-8}$.
View full question & answer→Question 681 Mark
Use the properties of exponents to verify that each statement is true. $25(5^{\text{n}-2})=5^{\text{n}}$
Answer$25(5^{\text{n}-2})=5^{\text{n}}$ $\text{LHS}=25(5^{\text{n}-2})=5^{\text{2}}(5^{\text{n}}+5^2)$ $=5^2\times5^\text{n}\times\frac{1}{5^2}=5^{\text{n}}=\text{RHS}$
View full question & answer→Question 691 Mark
By multiplying $(10)^5$ by $(10)^{-10}$ we get ________.
AnswerBy multiplying $(10)^5$ by $(10)^{-10}$ we get $(10)^{-5}$.
Solution:
Using law of exponents, $(a)^m× (a)^n= (a)^{m+n}$ [$\because$ a is non-zero integer]
Similarly,
$(10)^5× (10)^{-10}+ (10)^{5-10}$
$= (10)^{-5}$
Hence, $(10)^5× (10)^{-10}=(10)^{-5}$
View full question & answer→Question 701 Mark
$(-4)^{-4}× (4)^{-1}= (4)^5$
Answer$LHS = (-4)^{-4}× (4)^{-1}$
Using law of exponents, $a^{m}× a^n= (a)^{m+n}[\because$ a is non-zero integer$]$
$\therefore$ $(-4)^{-4}× (4)^{-1}= (4)^{-4}× (4)^{-1}$
$= (-4)^{-4-1}$
$= (-4)^{-5}$
$LHS ≠ RHS$
View full question & answer→Question 711 Mark
$5^5× 5^{-5}$= __________.
Answer$5^5× 5^{-5}= 1.$
Solution:
Using law of exponents, $a^m× a^n= (a)^{m+n}$[$\because$ a is non-zero integer]
$\therefore$ $5^5× 5^{-5}= (5)^{5-5}$
$= (5)^0= 1$ [$\therefore$ $a^0= 1$]
Hence,
$5^5× 5^{-5}= 1$
View full question & answer→Question 721 Mark
The standard form for $0.000037$ is $3.7 × 10^{-5}$.
AnswerFor standard form,$ 0.000037 = 0.37 × 10^{-4}$
$= 3.7 × 10^{-5}$
View full question & answer→Question 731 Mark
The volume of the Earth is approximately $7.67 × 10^{-7}$ times the volume of the Sun. Express this figure in usual form.
AnswerGiven, volume of the Earth is $7.67 × 10^{-7}$ times the volume of the sun Usual form of $7.67 × 10^{-7}= 0.000000767.$ $[\because$ placing decimal $7$ places towards the left of original position$]$
View full question & answer→Question 741 Mark
$5^0= 5$
Answer$LHS = 5^0$
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore 5^0= 1$
$LHS ≠ RHS$
View full question & answer→Question 751 Mark
$a^3 \times a^{-10}=$= __________.
Answer$a^3 \times a^{-10}= a^{-7}.$
Solution:
Given,
Using law of exponents, $a^m \times a^n=(a)^{m+n}$[$\because$ a is non-zero integer]
Similarly,
$a^3 \times a^{-10}=(a)^{3-10}$
$=(a)^{-7}$
Hence,
$a^3 \times a^{-10}=a^{-7}$
View full question & answer→Question 761 Mark
$(-7)^{-4} \times(-7)^2=(-7)^{-2}$
Answer$LHS =(-7)^{-4} \times(-7)^2$
Using law of exponents, $a^m \times a^n=(a)^{m+n}$ [$\because$ a is non-zero integer]
$\therefore(-7)^{-4} \times(-7)^2=(-7)^{-4+2}$
$=(-7)^{-2}$
$LHS ≠ RHS$
View full question & answer→Question 771 Mark
$(-6)^0= -1$
Answer$LHS = (-6)^0$
Using law of exponents, $a^0= 1 [ \because$ a is non-zero integer$]$
$\therefore (-6)^0= -1$
$LHS ≠ RHS$
View full question & answer→Question 781 Mark
The usual form of $2.39461 × 10^6$ is ________.
AnswerThe usual form of $2.39461 \times 10^6$ is $2394610.$
Solution:
For usual form, $2.39461 \times 10^6= 2394610 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$
$= 2394610$
Hence,
The usual form of $2.39461 \times 10^6$ is $2394610.$
View full question & answer→Question 791 Mark
On dividing $8^5$ by _________ we get $8.$
AnswerLet $8^5$ be divided by $x$ to get $8.$
So, $8^5÷ x = 8$
$\Rightarrow8^5\times\frac{1}{\text{x}}=8$
$\Rightarrow\frac{8^5}{8}=\text{x}$
$\therefore$ $\text{x}=\frac{8^5}{8^1}=8^{5-1}=8^4[\because$ $a^m÷ a^n= (a)^{m-n}]$
View full question & answer→Question 801 Mark
$329.25=3 \times 10^2+2 \times 10^1+9 \times 10^0+2 \times 10^{-1}+5 \times 10^{-2}$
Answer$RHS =3 \times 10^2+2 \times 10^1+9 \times 10^0+2 \times 10^{-1}+5 \times 10^{-2}$
$=3\times10\times10+2\times10+9\times1+\frac{2}{10}+\frac{5}{10\times10}[\because a^0= 1] $
$=300+20+9+0.2-0.05=329.25$
$\therefore LHS = RHS$
View full question & answer→Question 811 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $ (a^m× a^n= a^{m+n})$ $[\therefore$ a is non-zero integer$]$
Repeator machine can do the work is equal to $2^2\times\big(\frac{1}{2}\big)^3\times2^4=2^6\times\frac{1}{2^3}=2^3$$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m}-\text{n}}\Big]$
so, $(x2^3)$ single machine can do the same work.
View full question & answer→Question 821 Mark
Very large numbers can be expressed in standard form by using _________ exponents.
AnswerVery large numbers can be expressed in standard form by using positive exponents.
Solution:
Very large numbers can be expressed in standard form by using positive exponents,
$ 23000=23 \times 10^3$
$ =2.3 \times 10^3 \times 10^1 $
$=2.3 \times 10^4 $
View full question & answer→Question 831 Mark
The standard form for $0.000000008$ is __________.
AnswerThe standard form for $0.000000008$ is $8.0 × 10^{-9}$.
Soluiton:
For standard form, $0.000000008 = 0.8 \times 10^{-8}= 8 \times 10^{-9}= 8.0 \times 10^{-9}$
Hence,
The standard form for $0.000000008$ is $8.0\times 10^{-9}$
View full question & answer→Question 841 Mark
Find the value $ \left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
AnswerUsing law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}[ \because$ a is non-zero integer$]$
$\therefore\left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
$=\Big(\frac{1}{4}+\frac{1}{3}+\frac{1}{36}\Big)^{-1}$
$=\Big(\frac{9+12+1}{36}\Big)^{-1}[\because LCM$ of $4, 3,$ and $36 = 36]$
$=\Big(\frac{22}{36}\Big)^{-1}=\frac{36}{22}=\frac{18}{11}$
Hence,
$\Big[4^{-1}+3^{-1}+6^{-2}\Big]^{-1}=\frac{18}{11}$
View full question & answer→Question 851 Mark
$a^m× b^m= (ab)^m$
Answer$LHS = a^m× b^m= (a × b)^m= (ab)^m$
$[$by law of exponents$]$
View full question & answer→Question 861 Mark
Supply the missing information for diagram.
AnswerIf $1.25\ cm$ long piece is inserted in $(x4)$ repeated machine, then it will produce $1.25 × 4 = 10\ cm$ long piece. $\because \ ?=10\text{cm}$
View full question & answer→Question 871 Mark
Large numbers can be expressed in the standard form by using positive exponents.
Answere.g. $2360000 = 236 \times 10 \times 10 \times 10 \times 10$
$ =236 \times 10^4 $
$ =2.36 \times 10^4 \times 10^2 $
$ =2.36 \times 10^6 $
View full question & answer→Question 881 Mark
The multiplicative inverse of $(-4)^{-2}$ is $(4)^{-2}$.
Answer$a$ is called the multiplicative inverse of b, if $a \times b = 1.$
Put $b = (-4)^{-2}$
$\therefore$ $a × (-4)^{-2}= 1$
$\Rightarrow\text{a}=\frac{1}{(-4)^{-2}}=(-4)^2$
View full question & answer→Question 891 Mark
$\frac{\text{x}^\text{m}}{\text{y}^\text{m}}=\Big(\frac{\text{y}}{\text{x}}\Big)^{-\text{m}}$
Answer$RHS =\Big(\frac{\text{y}}{\text{x}}\Big)^{-\text{m}}$
Using law of exponents, $\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}}=\frac{\text{a}^\text{m}}{\text{b}^\text{m}}$ and $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}} [\because$ a is non-zero integers$]$
$\therefore \Big(\frac{\text{y}}{\text{x}}\Big)^{-\text{m}}=\frac{\text{y}^{-\text{m}}}{\text{y}^{-\text{m}}}$
$=\frac{\text{x}^\text{m}}{\text{y}^\text{m}} $
$LHS = RHS$
View full question & answer→Question 901 Mark
$ 3^5 \div 3^{-6}$ can be simplified as __________.
Answer$3^5 \div 3^{-6}$ can be simplified as $(3)^{11}$.
Solution:
Given,
$3^5 \div 3^{-6}=(3)^{5-(-6)}$
$=(3)^{5+6}=(3)^{11}\left[\because a^m+a^n=(a)^{m-n}\right]$
Hence,
$3^5 \div 3^{-6}$ can be simplified as $(3)^{11}$.
View full question & answer→Question 911 Mark
The expression for $8^{-2}$ as a power with the base $2$ is _________.
AnswerGiven,
$8^{-2}$, Where we can write $8 = 2 \times 2 \times 2$
$\therefore(2 \times 2 \times 2)^{-2}=(2)^{3 \times(-2)}=(2)^{-6}$
Hence,
$8^{-2}=(2)^{-6}$
View full question & answer→Question 921 Mark
$(-2)^0= 2$
Answer$LHS = (-2)^0$
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore (-2)^0= 1$
$LHS \neq RHS$
View full question & answer→Question 931 Mark
If $36 = 6 × 6 = 6^2$, then $\frac{1}{36}$ expressed as a power with the base $6$ is ________.
AnswerGiven,
$36 = 6 \times 6 = 6^2$
So,
$\frac{1}{36}=\frac{1}{6\times6}=\frac{1}{(6)^2}=(6)^{-2}$
$\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→Question 941 Mark
$\text{a}^\text{m}=\frac{1}{\text{a}^{-\text{m}}}$
AnswerUsing law of exponents, $\text{a}^\text{m}=\frac{1}{\text{a}^{-\text{m}}}$
$LHS = RHS$
View full question & answer→Question 951 Mark
The multiplicative inverse of $10^{10}$ is ___________.
AnswerFor multiplicative inverse, a is called the multiplicative inverse of $b,$ if $a \times b = 1$
Put $b = 10^{10}$
Then,
$a \times 10^{10}= 1$
$\Rightarrow\text{a}=\frac{1}{10}^{10}\ \Big[\because\frac{1}{\text{a}^\text{m}}=\text{a}^{-\text{m}}\Big]$
$\therefore\text{a}=10^{-10}$
Hence,
The multiplicative inverse of $10^{10}$ is $10^{-10}$.
View full question & answer→