If the point (3, 4) lies on the graph of
Question: If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. Solution: Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 = 12 = 3a+7 = 3a = 12 7 = 3a = 5 Hence, the value of a is 5/3....
Read More →Find the following product:
Question: Find the following product:4.1xy(1.1xy) Solution: To find the product, we will use distributive law as follows: 4. $1 x y(1.1 x-y)$ $=(4.1 x y \times 1.1 x)-(4.1 x y \times y)$ $=\{(4.1 \times 1.1) \times x y \times x\}-(4.1 x y \times y)$ $=\left(4.51 x^{1+1} y\right)-\left(4.1 x y^{1+1}\right)$ $=4.51 x^{2} y-4.1 x y^{2}$ Thus, the answer is $4.51 x^{2} y-4.1 x y^{2}$....
Read More →Write the linear equation such that
Question: Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. Solution: Let the abscissa of the point be x,According to the question, Ordinate (y) = 3 x Abscissa = y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, we find two points $A(1,3)$ and $B(2,6)$. So, draw the graph by plotting the points and joining the line $A B$. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its...
Read More →From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°.
Question: From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30 and 45. The height of the hill is (a) $\frac{1}{2}(\sqrt{3}-1) \mathrm{km}$ (b) $\frac{1}{2}(\sqrt{3}+1) \mathrm{km}$ (c) $(\sqrt{3}-1) \mathrm{km}$ (d) $(\sqrt{3}+1) \mathrm{km}$ Solution: (b) $\frac{1}{2}(\sqrt{3}+1) \mathrm{km}$ Let $A B$ be the hill making angles of depression at points $C$ and $D$ such that $\angle A D B=45^{\circ}, \angle A C B=30^{\circ}$ and $C D=1 \mathrm{...
Read More →Find the following product:
Question: Find the following product: 1.5x(10x2y 100xy2) Solution: To find the product, we will use distributive law as follows: 1. $5 x\left(10 x^{2} y-100 x y^{2}\right)$\ $=\left(1.5 x \times 10 x^{2} y\right)-\left(1.5 x \times 100 x y^{2}\right)$ $=\left(15 x^{1+2} y\right)-\left(150 x^{1+1} y^{2}\right)$ $=15 x^{3} y-150 x^{2} y^{2}$ Thus, the answer is $15 x^{3} y-150 x^{2} y^{2}$....
Read More →Draw the graph of the linear equation
Question: Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. Solution: As per question, the sum of the coordinates is 10 units. Let x and y be two coordinates, then we get x + y = 10. For x = 5, y = 5, therefore, (5, 5) lies on the graph of x + y = 10. For x = 3, y = 7, therefore, (3, 7) lies on the graph of x + y = 10. Now, plotting the points (5, 5) and (3, 7) on the graph paper and joining them by a line, we get g...
Read More →Find the following product:
Question: Find the following product: $-\frac{4}{27} x y z\left(\frac{9}{2} x^{2} y z-\frac{3}{4} x y z^{2}\right)$ Solution: To find the product, we will use distributive law as follows: $-\frac{4}{27} x y z\left(\frac{9}{2} x^{2} y z-\frac{3}{4} x y z^{2}\right)$ $=\left\{\left(-\frac{4}{27} x y z\right)\left(\frac{9}{2} x^{2} y z\right)\right\}-\left\{\left(-\frac{4}{27} x y z\right)\left(\frac{3}{4} x y z^{2}\right)\right\}$ $=\left\{\left(-\frac{4}{27} \times \frac{9}{2}\right)\left(x^{1+2}...
Read More →Draw the graph of the equation represented
Question: Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. Solution: Any straight line parallel to X-axis in negative direction of Y-axis is given by y = k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To draw the graph of this equation, plot the points (1,3), (2, -3) and (3, -3) and join them. This is the required graph....
Read More →Determine the point on the graph of the linear equation 2x+ 5y = 19
Question: Determine the point on the graph of the linear equation 2x+ 5y = 19 whose ordinate is 1 times its abscissa. Thinking Process (i)Firstly, consider abscissa as x and ordinate as y and make a linear equation under the given condition. (ii)Solving both linear equations to get the value of x and y. (iii)Further, write the coordinates in a point form. Solution: Let $x$ be the abscissa of the given line $2 x+5 y=19$, then by given condition, Ordinate $(y)=1 \frac{1}{2} \times$ Abscissa $\Righ...
Read More →Find the following product:
Question: Find the following product: $-\frac{8}{27} x y z\left(\frac{3}{2} x y z^{2}-\frac{9}{4} x y^{2} z^{3}\right)$ Solution: To find the product, we will use the distributive law in the following way: $-\frac{8}{27} x y z\left(\frac{3}{2} x y z^{2}-\frac{9}{4} x y^{2} z^{3}\right)$ $=\left\{\left(-\frac{8}{27} x y z\right)\left(\frac{3}{2} x y z^{2}\right)\right\}-\left\{\left(-\frac{8}{27} x y z\right)\left(\frac{9}{4} x y^{2} z^{3}\right)\right\}$ $=\left\{\left(-\frac{8}{27} \times \frac...
Read More →Draw the graphs of linear equations
Question: Draw the graphs of linear equations y = x and y = x on the same Cartesian plane. What do you observe? f Thinking Process (i)Firstly find atleast two different points satisfying each linear equation. (ii)Secondly plot these points on a graph paper and get two different lines respective after joining their points. (iii)Further, observe the equations of lines. Solution: The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. Fo...
Read More →In a rectangle, the angle between a diagonal and a side is 30°
Question: In a rectangle, the angle between a diagonal and a side is 30 and the length of this diagonal is 8 cm. the area of the rectangle is (a) $16 \mathrm{~cm}^{2}$ (b) $\frac{16}{\sqrt{3}} \mathrm{~cm}^{2}$ (c) $16 \sqrt{3} \mathrm{~cm}^{2}$ (d) $8 \sqrt{3} \mathrm{~cm}^{2}$ Solution: (c) $16 \sqrt{3} \mathrm{~cm}^{2}$ Let $A B C D$ be the rectangle in which $\angle B A C=30^{\circ}$ and $A C=8 \mathrm{~cm}$. In $\Delta B A C$, we have: $\frac{A B}{A C}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$ $\...
Read More →Find the following product:
Question: Find the following product: $\left(-\frac{7}{4} a b^{2} c-\frac{6}{25} a^{2} c^{2}\right)\left(-50 a^{2} b^{2} c^{2}\right)$ Solution: To find the product, we will use distributive law as follows: $\left(-\frac{7}{4} a b^{2} c-\frac{6}{25} a^{2} c^{2}\right)\left(-50 a^{2} b^{2} c^{2}\right)$ $=\left\{\left(-\frac{7}{4} a b^{2} c\right)\left(-50 a^{2} b^{2} c^{2}\right)\right\}-\left\{\left(\frac{6}{25} a^{2} c^{2}\right)\left(-50 a^{2} b^{2} c^{2}\right)\right\}$ $=\left\{\left\{-\fra...
Read More →The graph of every linear equation in
Question: The graph of every linear equation in two variables need not be a line. Solution: FalseSince, the graph of a linear equation in two variables always represent a line....
Read More →Every point on the graph of a linear
Question: Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation. Solution: FalseSince, every point on the graph of the linear equation represents a solution....
Read More →The coordinates of points in the table
Question: The coordinates of points in the table represent some of the solutions of the equation x y + 2 = 0. Solution: False The coordinates of points are (0, 2), (1,3), (2, 4), (3, 5) and (4, 6). Given equation is x-y+2 = 0 At point (0,2), 0 2 + 2 = 0 = 0=0, it satisfies. At point (1,3), 1-3+2 = 3-3 = 0 = 0 = 0, it satisfies. At point (2, 4), 2-4+2 = 4- 4 = 0 = 0 = 0, it satisfies. At point (3,-5), 3 (- 5) + 2 = 3 + 5 + 2 = 10 0, it does not satisfy. At point (4, 6), 4-6+2-6-6 = 0 = 0 = 0, it ...
Read More →Find the following product:
Question: Find the following product:0.1y(0.1x5+ 0.1y) Solution: To find the product, we will use distributive law as follows: $0.1 y\left(0.1 x^{5}+0.1 y\right)$ $=(0.1 y)\left(0.1 x^{5}\right)+(0.1 y)(0.1 y)$ $=(0.1 \times 0.1)\left(y \times x^{5}\right)+(0.1 \times 0.1)(y \times y)$ $=(0.1 \times 0.1)\left(x^{5} \times y\right)+(0.1 \times 0.1)\left(y^{1+1}\right)$ $=0.01 x^{5} y+0.01 y^{2}$ Thus, the answer is $0.01 x^{5} y+0.01 y^{2}$....
Read More →On the level ground, the angle of elevation of a tower is 30°.
Question: On the level ground, the angle of elevation of a tower is 30. On moving 20 m nearer, the angle of elevation is 60. The height of the tower is(a) 10 m (b) $10 \sqrt{3} \mathrm{~m}$ (c) $15 \mathrm{~m}$ (d) $5 \sqrt{3} \mathrm{~m}$ Solution: (b) $10 \sqrt{3} \mathrm{~m}$ Let $A B$ be the tower and $C$ and $D$ be the points of observation such that $\angle B C D=30^{\circ}, \angle B D A=60^{\circ}, C D=20 \mathrm{~m}$ and $A D=x \mathrm{~m}$. Now, in $\triangle A D B$, we have: $\frac{A B...
Read More →The graph given below represents the linear equation x = 3.
Question: The graph given below represents the linear equation x = 3. Solution: True Since, given graph is a line parallel to y-axis at a distance 3 units to the right of the origin. Hence, it represents a linear equation x = 3....
Read More →Solve the following
Question: xy(x3y3) Solution: To find the product, we will use the distributive law in the following way: $x y\left(x^{3}-y^{3}\right)$ $=x y \times x^{3}-x y \times y^{3}$ $=\left(x \times x^{3}\right) \times y-x \times\left(y \times y^{3}\right)$ $=x^{1+3} y-x y^{1+3}$ $=x^{4} y-x y^{4}$ Thus, the answer is $x^{4} y-x y^{4}$....
Read More →The graph given below represents the linear equation x + y = 0.
Question: The graph given below represents the linear equation x + y = 0. Solution: True If the given points (-1,1) and (- 3, 3) lie on the linear equation x + y = 0, then both points will satisfy the equation. So, at point (-1,1), we put x = -1, and y = 1 in LHS of the given equation, we get LHS = x + y = -1+1 = 0 = RHS Again, at point (-3 3) put x = 3 and y = 3 in LHS of the given equation, we get LHS = x+ y= 3+ 3= 0 = RHS Hence, (-1,1) and (-3, 3) both satisfy the given linear equation x + y ...
Read More →Find the following product:
Question: Find the following product: $\frac{6 x}{5}\left(x^{3}+y^{3}\right)$ Solution: To find the product, we will use distributive law as follows: $\frac{6 x}{5}\left(x^{3}+y^{3}\right)$ $=\frac{6 x}{5} \times x^{3}+\frac{6 x}{5} \times y^{3}$ $=\frac{6}{5} \times\left(x \times x^{3}\right)+\frac{6}{5} \times\left(x \times y^{3}\right)$ $=\frac{6}{5} \times\left(x^{1+3}\right)+\frac{6}{5} \times\left(x \times y^{3}\right)$ $=\frac{6 x^{4}}{5}+\frac{6 x y^{3}}{5}$ Thus, the answer is $\frac{6 ...
Read More →The graph of the linear equation x + 2y = 7
Question: The graph of the linear equation x + 2y = 7 passes through the point (0, 7). Solution: False If we put x = 0 and y = 7 in LHS of the given equation, we get LHS = (0) + 2 (7)= 0 + 14 = 14 7 = RHS Hence, (0, 7) does not lie on the line x + 2y = 7....
Read More →Find the following product:
Question: Find the following product:11y2(3y + 7) Solution: To find the product, we will use distributive law as follows: $-11 y^{2}(3 y+7)$ $=\left(-11 y^{2}\right) \times 3 y+\left(-11 y^{2}\right) \times 7$ $=(-11 \times 3)\left(y^{2} \times y\right)+(-11 \times 7) \times\left(y^{2}\right)$ $=(-33)\left(y^{2+1}\right)+(-77) \times\left(y^{2}\right)$ $=-33 y^{3}-77 y^{2}$ Thus, the answer is $-33 y^{3}-77 y^{2}$....
Read More →The point (0, 3) lies on the graph
Question: The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12. Solution: True If we put x = 0 and y = 3 in LHS of the given equation, we find LHS = 3 x 0+4 x 3 = 0+12 = 12 = RHS Hence, (0, 3) lies on the linear equation 3x + 4y = 12....
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