The number log2 7 is : -Maths 9th

The number log2 7 is : -Maths 9th
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Answer :

(c) an irrational numberLet us assume log27 be a rational number. Then, log27 = \(rac{p}{q}\),  where p,q ∈ I and q ≠ 0 ⇒ \(2^{rac{p}{q}}\) = 7 ⇒ 2p = 7q This is not true as 2 is even and 7 is odd.∴ Hence our assumption that log27 is a rational number is wrong. ∴ log27 is an irrational number.
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Related questions

The number of meaningful solutions of log4(x – 1) = log2 (x – 3) is -Maths 9th

Description : The number of meaningful solutions of log4(x – 1) = log2 (x – 3) is -Maths 9th

Last Answer : (b) 1log4(x - 1) = log2(x - 3) ⇒ log22 (x − 1) = log2(x - 3)⇒ \(rac{1}{2}\) log2 (x-1) = log2 (x- 3) ⇒ log2 (x-1) = 2 log2 (x- 3)\(\big[\)Using logam (bn) = \(rac{n}{m} ... x = 2 or 5 Neglecting x = 2 as log2(x - 3) is defined when x > 2.⇒ There is only one meaningful solution of the given equation.

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Find the value of x, if log2 (5.2^x + 1), log4(2^(1–x) + 1) and 1 are in A.P. -Maths 9th

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Last Answer : (b) 1 - log25 Given, log2 (5.2x + 1), log4 (21- x + 1), 1 are in A.P. ⇒ log2 (5.2x + 1) + 1 = 2 log4 (21 - x + 1) ⇒ log2 (5.2x + 1) + log22 = 2 log22 (21-x + 1)⇒ log2 (5.2x + 1).2 = 2 x \(rac12\) ... a=-rac{1}{2}\big)\)⇒ log 2x = log \(rac{2}{5}\)⇒ x log2 2 = log2 2 - log2 5 ⇒ \(x\) = 1 - log2 5.

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Last Answer : (c) 4log2 [log7(x2 – x + 37)] = 1 ⇒ log7(x2 – x + 37) = 21 = 2 ⇒ x2 – x + 37 = 72 = 49 ⇒ x2 – x – 12 = 0 Now solve for x.

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Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

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A football player scored the following number of goals in the 10 matches 1, 3, 2, 5, 8, 6,1, 4, 7 and 9. Since, the number of matches is 10 (an even number), therefore -Maths 9th

Description : A football player scored the following number of goals in the 10 matches 1, 3, 2, 5, 8, 6,1, 4, 7 and 9. Since, the number of matches is 10 (an even number), therefore -Maths 9th

Last Answer : NEED ANSWER

A football player scored the following number of goals in the 10 matches 1, 3, 2, 5, 8, 6,1, 4, 7 and 9. Since, the number of matches is 10 (an even number), therefore -Maths 9th

Description : A football player scored the following number of goals in the 10 matches 1, 3, 2, 5, 8, 6,1, 4, 7 and 9. Since, the number of matches is 10 (an even number), therefore -Maths 9th

Last Answer : No. It is not the correct answer, because the data have to be arranged in ascending or descending order before finding the median. Arranging the data in ascending order 1,1,2, 3, 4, 5, 6, 7, 8, 9. Here, number of observations is 10, which is even.

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Description : If n(A) = 5 and n(B) = 7, then the number of relations on A × B is -Maths 9th

Last Answer : (b) 235n(A) = 5 and n(B) = 7 ∴ n(A × B) = 5 × 7 = 35 Total number of relations from A to B = Total number of subsets of (A × B) = 235.

Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

7. Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number? -Maths 9th

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Last Answer : (a) 0.8 lies between 0 and 1 0.8 is nearer to 1 (b) 5.1 lies between 5 and 6 5.1 is nearer to 5 (c) 2.6 lies between 2 and 3 2.6 is nearer to 3 (d) 6.4 lies between 6 and 7 6.4 is nearer to 6 (e) 9.1 lies between 9 and 10 9.1 is nearer to 9 (f) 4.9 lies between 4 and 5 4.9 is nearer to 5

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Description : 3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

Last Answer : Solution: 7+3x = 0 ⇒ 3x = −7 ⇒ x = -7/3 ∴Remainder: 3(-7/3)3+7(-7/3) = -(343/9)+(-49/3) = (-343-(49)3)/9 = (-343-147)/9 = -490/9 ≠ 0 ∴7+3x is not a factor of 3x3+7x

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Description : Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x2–3x+7 -Maths 9th

Last Answer : Solution: The equation 4x2–3x+7 can be written as 4x2–3x1+7x0 Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2–3x+7 is a polynomial in one variable

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Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

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The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Last Answer : Length (l) of room = 5m Breadth (b) of room = 4m Height (h) of room = 3m It can be observed that four walls and the ceiling of the room are to be white washed. Total area to be white washed = Area of walls + ... m2 area = Rs.7.50 (Given) Cost of white washing 74 m2 area = Rs. (74 7.50) = Rs. 555

Find two irrational numbers between ROOT 2 and ROOT 7. -Maths 9th

Description : Find two irrational numbers between ROOT 2 and ROOT 7. -Maths 9th

Last Answer : Root 3 root 5

Find the decimal expansion of 7/22 .Explain why it is taken as the approximate value of pi -Maths 9th

Description : Find the decimal expansion of 7/22 .Explain why it is taken as the approximate value of pi -Maths 9th

Last Answer : 3.1564267467648582474747477647908765434567896542345678765431222245775583585.................................. or 22/7 is the value f pi

Find the five rational no. Between5/7 and9/11 -Maths 9th

Description : Find the five rational no. Between5/7 and9/11 -Maths 9th

Last Answer : 5/7 and 9/11 find the lcm of 7 and 11 lcm =77. 5/7 *11/11=55/77 , 9/11*7/7=63/77 therefore five rational no are 56/77 57/77 58/77 59/77 60/77

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

Show that 0.142857142857… = 1/7. -Maths 9th

Description : Show that 0.142857142857… = 1/7. -Maths 9th

Last Answer : Let x = 0.142857142857 …………………..(i) On multiplying both sides of Eq. (i) by 1000000, we get 1000000 x = 142857.142857…………………………(ii) On subtracting Eq. (i) from Eq. (ii), we get 1000000 x – x = (142857.142857…) – (0.142857..) ⇒ 999999 x = 142857 ∴ x = 142857/999999 = 1/7 Hence proved.

Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

Point (0, – 7) lies -Maths 9th

Description : Point (0, – 7) lies -Maths 9th

Last Answer : (c) In point (0, -7) x-coordinate is zero, so it lies on Y-axis and y-coordinate is negative, so the point (0, – 7) lies on the Y-axis in the negative direction.

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

The linear equation 2x – 5y = 7 has -Maths 9th

Description : The linear equation 2x – 5y = 7 has -Maths 9th

Last Answer : (c) In the given equation 2x – 5y = 7, for every value of x, we get a corresponding value of y and vice-versa. Therefore, the linear equation has infinitely many solutions.

The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Last Answer : (a) In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.

The equation x = 7, in two variables can be written as -Maths 9th

Description : The equation x = 7, in two variables can be written as -Maths 9th

Last Answer : (b) Here, the’Coefficient of y in the given equation x =7 is 0. So, the equation can be written as 1-x + 0-y = 7 Hence, the required equation is 1.x + 0. y = 7.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : 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.

If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

Find two irrational numbers between ROOT 2 and ROOT 7. -Maths 9th

Description : Find two irrational numbers between ROOT 2 and ROOT 7. -Maths 9th

Last Answer : Root 3 root 5

Find the decimal expansion of 7/22 .Explain why it is taken as the approximate value of pi -Maths 9th

Description : Find the decimal expansion of 7/22 .Explain why it is taken as the approximate value of pi -Maths 9th

Last Answer : 3.1564267467648582474747477647908765434567896542345678765431222245775583585.................................. or 22/7 is the value f pi

Find the five rational no. Between5/7 and9/11 -Maths 9th

Description : Find the five rational no. Between5/7 and9/11 -Maths 9th

Last Answer : 5/7 and 9/11 find the lcm of 7 and 11 lcm =77. 5/7 *11/11=55/77 , 9/11*7/7=63/77 therefore five rational no are 56/77 57/77 58/77 59/77 60/77

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44