Find each of the following product:
Question: Find each of the following product: (4x2) (6xy2) (3yz2) Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $\left(-4 x^{2}\right) \times\left(-6 x y^{2}\right) \times\left(-3 y z^{2}\right)$ $=\{(-4) \times(-6) \times(-3)\} \times\left(x^{2} \times x\right) \times\left(y^{2} \times y\right) \times z^{2}$ $=\{(-4) \times(-6) \times(-3)\} \times\left(x^{2+1}\right) \times\left(y^...
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Question: $2 x-y=-2$ $3 x+4 y=3$ Solution: Given: $2 x-y=-2$ $3 x+4 y=3$ Using Cramer's Rule, we get $\mathrm{D}=\mid 2-1$ $3 \quad 4 \mid=8+3=11$ $\mathrm{D}_{1}=\mid-2-1$ $3 \quad 4 \mid=-8+3=-5$ $\mathrm{D}_{2}=\mid 2-2$ $3 \quad 3 \mid=6+6=12$ Now, $\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{-5}{11}$ $\mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}=\frac{12}{11}$ $\therefore \mathrm{x}=-\frac{5}{11}$ and $\mathrm{y}=\frac{12}{11}$...
Read More →The points whose abscissa and
Question: The points whose abscissa and ordinate have different signs will lie in (a)I and II quadrants (b)II and III quadrants (c)I and III quadrants (d)II and IV quadrants Solution: (d)The points whose abscissa and ordinate have different signs will be of the form (-x, y) or (x, y)and these points will lie in II and IV quadrants....
Read More →Find each of the following product:
Question: Find each of the following product: (5a) (10a2) (2a3) Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $(-5 a) \times\left(-10 a^{2}\right) \times\left(-2 a^{3}\right)$ $=\{(-5) \times(-10) \times(-2)\} \times\left(a \times a^{2} \times a^{3}\right)$ $=\{(-5) \times(-10) \times(-2)\} \times\left(a^{1+2+3}\right)$ $=\{(-5) \times(-10) \times(-2)\} \times\left(a^{1+2+3}\right)$...
Read More →Find each of the following product:
Question: Find each of the following product: (7ab)(5ab2c)(6abc2) Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $(7 a b) \times\left(-5 a b^{2} c\right) \times\left(6 a b c^{2}\right)$ $=\{7 \times(-5) \times 6\} \times(a \times a \times a) \times\left(b \times b^{2} \times b\right) \times\left(c \times c^{2}\right)$ $=\{7 \times(-5) \times 6\} \times\left(a^{1+1+1}\right) \times\le...
Read More →A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60° to 30°.
Question: A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60 to 30. Find the speed of the boat in metres per minute.[Use $\sqrt{3}=1.732 .]$ Solution: Let AB be the light house. Suppose C and D be the two positions of the boat.Here, AB = 100 m.Let the speed of the boat bevm/min.So,CD =vm/min 2 min = 2vm [Distance = Speed Time]In right ∆ABC, $\tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$ $\Rightarrow \...
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Question: Find each of the following product: $(-7 x y) \times\left(\frac{1}{4} x^{2} y z\right)$ Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $(-7 x y) \times\left(\frac{1}{4} x^{2} y z\right)$ $=\left(-7 \times \frac{1}{4}\right) \times\left(x \times x^{2}\right) \times(y \times y) \times z$ $=\left(-7 \times \frac{1}{4}\right) \times\left(x^{1+2}\right) \times\left(y^{1+1}\right...
Read More →Find each of the following product:
Question: Find each of the following product: $\left(-\frac{1}{27} a^{2} b^{2}\right) \times\left(\frac{9}{2} a^{3} b^{2} c^{2}\right)$ Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $\left(-\frac{1}{27} a^{2} b^{2}\right) \times\left(\frac{9}{2} a^{3} b^{2} c^{2}\right)$ $=\left(-\frac{1}{27} \times \frac{9}{2}\right) \times\left(a^{2} \times a^{3}\right) \times\left(b^{2} \times b^...
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Question: Find each of the following product: $\left(\frac{-24}{25} x^{3} z\right) \times\left(-\frac{15}{16} x z^{2} y\right)$ Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $\left(-\frac{24}{25} x^{3} z\right) \times\left(-\frac{15}{16} x z^{2} y\right)$ $=\left\{\left(-\frac{24}{25}\right) \times\left(-\frac{15}{16}\right)\right\} \times\left(x^{3} \times x\right) \times\left(z \t...
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Question: $3 x+y=19$ $3 x-y=23$ Solution: Given : $3 x+y=19$ $3 x-y=23$ Using Cramer's Rule, we get $\mathrm{D}=\mid 3 \quad 1$ $3-1 \mid=-3-3=-6$ $\mathrm{D}_{1}=\mid 19 \quad 1$ $23-1 \mid=-19-23=-42$ $\mathrm{D}_{2}=\mid 3 \quad 19$ $3 \quad 23 \mid=(3 \times 23)-(3 \times 19)=3 \times 4=12$ Now, $\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{-42}{-6}=7$ $\mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}=\frac{12}{-6}=-2$ $\therefore \mathrm{x}=7$ and $\mathrm{y}=-2$...
Read More →Find each of the following product:
Question: Find each of the following produ $\left(-\frac{7}{5} x y^{2} z\right) \times\left(\frac{13}{3} x^{2} y z^{2}\right)$ Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$. We have: $\left(-\frac{7}{5} x y^{2} z\right) \times\left(\frac{13}{3} x^{2} y z^{2}\right)$ $=\left(-\frac{7}{5} \times \frac{13}{3}\right) \times\left(x \times x^{2}\right) \times\left(y^{2} \times y\right) \times\left(...
Read More →If at some time of the day the ratio of the height of a vertically standing pole to the length of its shadow on the ground is
Question: If at some time of the day the ratio of the height of a vertically standing pole to the length of its shadow on the ground is $\sqrt{3}: 1$ then find the angle of elevation of the sun atthat time. Solution: Let AB be the vertically standing pole of heighthunits and CB be the length of its shadow ofs units. Since, the ratio of length of pole and its shadow at some time of day is given to be $\sqrt{3}: 1$. $\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\sqrt{3}}{1}$ $\therefore \ln \...
Read More →Abscissa of a point is positive in
Question: Abscissa of a point is positive in (a)I and II quadrants (b)I and IV quadrants (c)I quadrant (d)II quadrant Solution: (b)Abscissa of a point is positive in I and IV quadrants....
Read More →Find each of the following product:
Question: Find each of the following product: $\frac{1}{4} x y \times \frac{2}{3} x^{2} y z^{2}$ Solution: To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, $a^{m} \times a^{n}=a^{m+n}$. We have: $\frac{1}{4} x y \times \frac{2}{3} x^{2} y z^{2}$ $=\left(\frac{1}{4} \times \frac{2}{3}\right) \times\left(x \times x^{2}\right) \times(y \times y) \times z^{2}$ $=\left(\frac{1}{4} \times \frac{2}{3}\right) \times\left(x^{1+2}\right...
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Question: $2 x-y=17$ $3 x+5 y=6$ Solution: Given: $2 x-y=17$ $3 x+5 y=6$ Using Cramers Rule, we get $\mathrm{D}=\mid 2-1$ $3 \quad 5 \mid=10+3=13$ $\mathrm{D}_{1}=\mid 17-1$ $6 \quad 5 \mid=85+6=91$ $\mathrm{D}_{2}=\mid 2 \quad 17$ $3 \quad 6 \mid=12-51=-39$ Now, $\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{91}{13}=7$ $\mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}=\frac{-39}{13}=-3$ $\therefore \mathrm{x}=7$ and $\mathrm{y}=-3$...
Read More →From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45° and 30°
Question: From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45 and 30 respectively. Find the height of the hill. Solution: Let PQ be the hill of heighthkm. Let R and S be two consecutive kilometre stones, so the distance between them is 1 km.Let QR =x km. In $\Delta \mathrm{PQR}$, $\tan 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{QR}}$ $\Rightarrow 1=\frac{h}{x}$ $\Rightarrow h=x \quad \ldots(i)$ In $\triangle \mathrm{PQS}$, $\tan 30^{\circ}=...
Read More →If P(5,1), Q(8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper,
Question: If P(5,1), Q(8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points) on the X-axis is/are (a)P and R (b)R and S (c)Only Q (d)Q and O Solution: (d)We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis....
Read More →Find each of the following product:
Question: Find each of the following product:(5xy) (3x2yz) Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, $a^{m} \times a^{n}=a^{m+n}$, wherever applicable. We have: $(-5 x y) \times\left(-3 x^{2} y z\right)$ $=\{(-5) \times(-3)\} \times\left(x \times x^{2}\right) \times(y \times y) \times z$ $=15 \times\left(x^{1+2}\right) \times\left(y^{1+1}\right) \times z$ $=15 x^{3} y^{2} z$ Thus, the answer is $15 x^{3} y^{2} z$....
Read More →If the coordinates of the two points are
Question: If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) (Abscissa of Q) is (a)-5 (b)1 (c)-1 (d)-2 Solution: (b)We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) (Abscissa of Q) = 2 (-3) =-2 + 3=1....
Read More →If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4)
Question: If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are (a)P and (b)Q and R (c)Only S (d)P and R Solution: (b)In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant....
Read More →Find each of the following product:
Question: Find each of the following product: 3a2 4b4 Solution: To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^{m} \times a^{n}=a^{m+n}$, wherever applicable. We have: $-3 a^{2} \times 4 b^{4}$ $=(-3 \times 4) \times\left(a^{2} \times b^{4}\right)$ $=-12 a^{2} b^{4}$ Thus, the answer is $-12 a^{2} b^{4}$....
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Question: $2 x-y=1$ $7 x-2 y=-7$ Solution: Given : $2 x-y=1$ $7 x-2 y=-7$ Using Crammer's Rule, we get $\mathrm{D}=\mid 2 \quad-1$ $7-2 \mid=-4+7=3$ $\mathrm{D}_{1}=\mid 1=-1$ $-7-2 \mid=-2-7=-9$ $\mathrm{D}_{2}=\mid 2 \quad 1$ $7-7 \mid=-14-7=-21$ Now, $\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{-9}{3}=-3$ $\mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}=\frac{-21}{3}=-7$ $\therefore \mathrm{x}=-3$ and $\mathrm{y}=-7$...
Read More →From a point on the ground the angles of elevation of the bottom and top of a communication tower fixed on the top of a 20-m-high building are 45°
Question: From a point on the ground the angles of elevation of the bottom and top of a communication tower fixed on the top of a 20-m-high building are 45 and 60 respectively. Find the height of the tower$[$ Take $\sqrt{3}=1.732]$ Solution: Let BC be the 20 m high building and AB be the communication tower of heighthfixed on top of the building. Let D be a point on ground such that CD =x m and angles of elevation made from this point to top and bottom of tower are $45^{\circ}$ and $60^{\circ}$ ...
Read More →Find each of the following product:
Question: Find each of the following product: 5x2 4x3 Solution: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions. In the present problem, to perform the multiplication, we can proceed as follows: $5 x^{2} \times 4 x^{3}=(5 \times 4) \times\left(x^{2} \times x^{3}\right)$ $=20 x^{5} \quad\left(\because a^{m} \times a^{n}=a^{m+n}\right)$ Thus, the answer ...
Read More →On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4)
Question: On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. Which of the following figure is obtained? (a)Square (b)Rectangle (c)Trapezium (d)Rhombus Solution: (b)Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle....
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