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| 62301. |
Using Apolloneous theorem, let us solve some examples.; Dample 13 : In ΔABC, AD, BE, CF are themedians. Prove that,4(AD2 + BE2 + CF2) 3(AB2BC2 + AC2). |
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Answer» thnx |
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| 62302. |
2. Test whether the given numbers are perfect cubes or not. 72951680002700243 |
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Answer» 729=9^38000=20^3except these are not perfect cube |
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| 62303. |
Check whether the following numbers are perfect cubes or not.a. 6859b. 13824C. 78312 |
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Answer» a. yes it is perfect cube, cube root of 6859 is 19 as ones place have 9 so only possiblity that number is having 9 is ones place so it will be 19 by inspection. b. yes it is perfect cube, cube root of 13824 is 24 as ones place have 4 so only possiblity that number is having 4 or 8 is ones place so it will be 24 by inspection. c. 78312 is not perfect cube as there is no number which is multiplied three times gives 2. |
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| 62304. |
Test whether the given numbers are perfect cubes or not.(i) 243(ii) 516(iii) 729(iv) 8000(v) 2700 |
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| 62305. |
(i) m + 6 = 2m - 4(ii) 3y - 4 = 2y + 2(iii) 5b + 9 = 2b + 3(iv) 3X-7 = 2x + 1(v) 5(n + 2) = 7n + 4 |
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Answer» M=10Y=6B= - 3X=8N=3 is correct answer for all questions 1)m+6=2m-4 6+4=2m-mm=102) 3y-4=2y+23y-2y=2+4y=6 (i) 10(ii) 6(iii) -2(iv) 8(v) 3 i) m=10ii)y= 6iii)b=-2Iv)X=8v)n=10 1. m=102. y=63. b=-24. x= 85. n=3is the correct answer me to 1)m+6=2m-46+4=2m-m10=m (1) m + 6 = 2m - 4 2m - m = 6 + 4 m = 10 (2) 3y - 4 = 2y + 2 3y - 2y = 2 + 4 y = 6 (3) 5b + 9 = 2b + 3 5b - 2b = 3 - 9 3b = -6 b = -6/3 b = -2 5)5(n+2)=7n+45n+10=7n+45n-7n=4-10-2n=-6n=3 n=3 is the right answer (i) m+6=2m-4 =6+4=2m-m =10=m =m= 10 1. m=102. y= 63. b= -24. X= 85. n= 3 (i) m + 6 = 2m - 4 4 + 6 = 2m - m 10 = m(ii) 3y - 4 = 2y + 2 3y - 2y = 2 + 4 y = 6(iii) 5b + 9 = 2b + 3 5b - 2b = 3 - 9 3b = -6 b = -6 ÷ 3 b = -2(iv) 3x - 7 = 2x + 1 3x - 2x = 1 + 7 x = 8(v) 5(n + 2) = 7n + 4(5 × n) + (5 × 2) = 7n + 4 5n + 10 = 7n + 4 10 - 4 = 7n - 5n 6 = 2n 6 ÷ 2 = n3 = n m=10y=6b=-2x=8n=3 (1) m-2m= -4-6 -m= -10 m=10(2) 3y-2y= 2+4 y=6(3) 5b-2b= 3-9 b=-6(4) 3x-2x= 1+7 x=8(5) 5n+10= 7n+4 5n-7n= 4-10 -2n= -6 n= -6/-2 n= 3 (I)m+6=2m-4 6+4=2m-mm=10(ii)3y-2y=2+4y=6(iii)5b-2b=3-93b=-6b=-2(iv)3x-2x=1+7x=8(v)5n+10=7n+410-4=7n-5n6=2nn=3 m=10y=6b=–6x=8n=3 is the best answer |
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| 62306. |
Prove that (2) :n-n-17n |
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| 62307. |
If the sum of 'n'terms of a series is 2n ^2+7n,Then find the first five terms of the progression. |
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| 62308. |
38. The ratio between the sum of n terms of two airthmetic progressions is (7n+ 1):(4n+2Find the ratio of their 11th terms |
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Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)We can consider the 11th term as the m th term.Let’s consider the ratio these two AP’s m th terms as am: a’m→(2)Recall the nth term of AP formula, an= a + (n – 1)dHence equation (2) becomes,am: a’m= a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].now substitute the value of m as 11so the answer becomes[ 14*11 - 6] : [8*11 + 23]= [ 148] : [111]= 4/3 = 4:3 |
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| 62309. |
3 Find the cube root of each of the following perfect cubes by estimationa 17576 b 132651 e 830584 0 21952 |
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Answer» a) 26 is the answer.b) 29 is the answer.c) 04 is the answer.d) 11 is the answer. |
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| 62310. |
est whether the given numbers are perfect cubes u(i) 243(ii) 516(v)2700(Gil) 729Find the smallest number by which 8788 must be multiplied to oblain(iv) 80003.muliplied with so that the product be |
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Answer» 1) 243 - no2) 516- no3) 729- yes.. It is a cube of 94) 800- yes.. It is a cube of 205) 2700- no |
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| 62311. |
3. 21written in lowest term isa)b)c)42 |
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Answer» 3. 7/21 written in lowest term is 1/3 7/21 written in lowest term is 1/3 |
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| 62312. |
10. Which of the following is in the lowest term?16128 |
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Answer» A. 16/128=0.125B. 4/24=1/6=0.1666/7=0.857D. 12/123=0.09 D. 12/123 is the lowest term lowest term is 12/123 |
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| 62313. |
Express each of the following as a fraction in its lowest term.a. 20% b. 8%. 82% |
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Answer» 20/100=1/58/100=2/2582/100=41/5067/100 1/5,2/25,41/59,67/100 is the best answer 20/100=1/58/100=2/2582/100=41/5067/100 20/100=1/58/100=2/2582/100=41/50 a) 20/100=1/5b) 8/100=2/25c) 82/100=41/50d) 67/100=67/100 |
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| 62314. |
4. How many three-digit numbers can beformed from the digits 0, 1, 3, 5, 6 ifrepetitions of digits (i) are allowed, (ii) arenot allowed? |
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Answer» 135,315,516,613, ya allows 135 156 635 316 536 356 013/135/356/560/610/and many more 013/135/356/560/610and many more right answer |
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| 62315. |
Let us apptble 13 : Hanumappa and his wife Gangamma areusy making jaggery out of sugarcane juice. They haveessed the sugarcane juice to make the molasses,pnocwhich is poured into moulds in the shape of a frustum ofa cone having the diameters of its two circular faces as0 cm and 35 cm and the vertical height of the mould is14 cm (see Fig.mass about 1.2 g, find the mass of the molasses that can3.22). If each cm of molasses has |
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| 62316. |
Sample 13: Hanumappa and his wife Gangamma aremaking jaggery out of sugarcane juice. They haverocessed the sugarcane juice to make the molasses,hich is poured into moulds in the shape of a frustum ofacone having the diameters of its two circular faces as30 cm and 35 cm and the vertical height of the mould is14 cm (see Fig. 13.22). If each cm of molasses hasmass about 1.2 g, find the mass of the molasses that canFig. 13.22 |
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Answer» Please like the solution 👍 ✔️ Given, mould is in the shape of a frustum of a cone in which r2= 30/2, r1=35/2 cm, h= 14 cm. Quantity (volume) molasses that can be poured into it= volume of the frustum = π/3 × h (r1² + r2²+r1r2)= 1/3×(22/7) × 14 [(35/2)²+(30/2)²+(35/2 × 30/2]= 44/3[ 306.25 + 225 +262.5]= 44/3 × 793.75= 11641.7 cm³ Given : 1 cm³ of molasses = 1.2 g Mass of the molasses that can be poured into each mould= 11641.7 × 1 2 = 13970.04 g= 13970.04/1000= 13.97 kg [ 1g = 1/1000 kg] Hence, Mass of the molasses that can be poured into each mould= 13.97 kg= 14 kg (approx) |
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| 62317. |
et us apply these formulae in some examplesmappa and his wife Gangamma areThey havebusy making jaggery out of sugarcane juice.processed the sugarcane juice to make the, molasses,which is poured into.moulds in the shape of a frustuma cone having the diameters of its two circular faces as30 cm and 35 cm and the vertical height of the mould is14 cm (see Fig. 13.22). If each cm2 of molasses hasmass about 1.2 g, find the mass of the molasses that canofFig. 13.22 |
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Answer» 1 2 3 |
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| 62318. |
Example 13: Hanumappa and his wibusy making jaggery out of sugarcane juice. They havefe Gangamma areprocessed the sugarcane juice to make the molasseswhich is poured into moulds in the shape of a frustum ofa cone having the diameters of its two circular faces as30 cm and 35 cm and the vertical height of the mould is14 cm (see Fig. 13.22). If each cm3 of molasses hasmass about 1.2 g, find the mass of the molasses that can |
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Answer» Please like the solution 👍 ✔️ Given, mould is in the shape of a frustum of a cone in which r2= 30/2, r1=35/2 cm, h= 14 cm. Quantity (volume) molasses that can be poured into it= volume of the frustum = π/3 × h (r1² + r2²+r1r2)= 1/3×(22/7) × 14 [(35/2)²+(30/2)²+(35/2 × 30/2]= 44/3[ 306.25 + 225 +262.5]= 44/3 × 793.75= 11641.7 cm³ Given : 1 cm³ of molasses = 1.2 g Mass of the molasses that can be poured into each mould= 11641.7 × 1 2 = 13970.04 g= 13970.04/1000= 13.97 kg [ 1g = 1/1000 kg] Hence, Mass of the molasses that can be poured into each mould= 13.97 kg= 14 kg (approx) |
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| 62319. |
^ { n } P _ { 4 } = 42 ^ { n } P _ { 2 } |
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Answer» how does(n-4) become (n-3) (n-2)!=(n-2)(n-3)(n-4)! |
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| 62320. |
7n23b342( p 5 |
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| 62321. |
Marked Price = 3600, Discount = 10% |
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Answer» The right answer is Rs 60.... sp= rs 540 is the correct answer of the given question right answer. is = 600× 10% = 60 the correct answer is 60 60rs is correct answer of following question is 60 is the answer of the following the correct answer is 60 60 is right answer**** market price = ₹ 600Discount = 10%=600 × 10/100=60₹selling price = 600 - 60 = 540₹. the correct answer is 60 600 rupees ka 10% =60,then600-60=540 answer the right answer is Rs 60 60 is the right answer........... 60 is the right answer |
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| 62322. |
1. Which of the following numbers are not perfect cubes?0 216(i) 128(ii) 1000(iv) 100(v) 46656 |
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Answer» all are perfect cubes except (ii)128, (iv)100 1,3,5 are perfect cube 1 3 are the perfect cubes all are perfect cubes excrpt 2)1284) 100 as far as I know and understand I didn't do anything wrong that my answer was marked as a spam. I just want to know why it happened? |
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| 62323. |
56. An item is sold at 5,225 after allowing a discount of 5%. Find its marked price. |
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Answer» Marked price= x-5*x/100= 5,225x-x/20= 522519x/20= 5225x= 5225*20/19 x= 5500 |
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| 62324. |
(0 720/ + 5)0/ - बाज 3600 ~ 4) (d) |
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| 62325. |
find the lowest term of 3600/7282 |
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| 62326. |
is marked at 3000. If the selling price of that item is 2700, find thediscount percentage. |
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Answer» 10'/. is the right answer |
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| 62327. |
6e there are successive discounts of10% and 20% on the following items. Findthe marked price of each item when theselling price is given.a) mobile phone 9,000 |
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Answer» market price are 10,000...... 10 thousands is the right answersand 20,000 1000 is the right answer 10,000 is right answer according to the given condition 10,000 is the correct answer 10,000 is the correct answer of the given question |
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| 62328. |
A shopkeeper fixed the marked price ofan item 35% above its cost price. Thepercentage ofdiscount allowed to gain 8% |
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| 62329. |
\begin{array}{l}{x-3 y-3=0} \\ {3 x-9 y-2=0} \\ {3 x-5 y=20} \\ {6 x-10 y=40}\end{array} |
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| 62330. |
xample 13 : Hanumappa and his wife Gangamma areusy making jaggery out of sugarcane juice. They havetocessed the sugarcane juice to make the molasseshich is poured into moulds in the shape of a frustum ofcone having the diameters of its two circular faces asDcm and 35 cm and the vertical height of the mould is4 cm (see Fig. 13.22). If each cm3 of molasses haslass about 1.2g, find the mass of the molasses that ca |
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Answer» Given, mould is in the shape of a frustum of a cone in which r2= 30/2, r1=35/2 cm, h= 14 cm. Quantity (volume) molasses that can be poured into it= volume of the frustum = π/3 × h (r1² + r2²+r1r2)= 1/3×(22/7) × 14 [(35/2)²+(30/2)²+(35/2 × 30/2]= 44/3[ 306.25 + 225 +262.5]= 44/3 × 793.75= 11641.7 cm³ Given : 1 cm³ of molasses = 1.2 g Mass of the molasses that can be poured into each mould= 11641.7 × 1 2 = 13970.04 g= 13970.04/1000= 13.97 kg [ 1g = 1/1000 kg] Hence, Mass of the molasses that can be poured into each mould= 13.97 kg= 14 kg (approx) Please like the solution 👍 ✔️ |
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| 62331. |
h of the equal0. The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the eousides. Find the length of each side of the triangle, area of the triangle and the heightofthetriangle. |
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Answer» Let the side of the isosceles triangle be xGIVEN3/2 x +x +x=42cm3x/2 + 2x =42(3x+4x)/2=423x+4x=847x=84x=12cmside=12cm base=18cmThe height(altitude) of an isos. trangle divides it into two rt tranglesin a rt.trangle, base²+heigrt²=hypotenuse²thus, 9²+h²=12²h=√12²-9²=√144-81 =√63 =3√7cmarea =1/2 x base x hieght =1/2 x 18 x3√7 27√7cm² |
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| 62332. |
h)2(p35)-(42),2p-5(p-29 |
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| 62333. |
\left. \begin{array} { l } { x - 3 y - 3 = 0 } \\ { 3 x - 9 y - 2 = 0 } \\ { 3 x - 5 y - 20 } \\ { 6 x - 10 y = 40 } \end{array} \right. |
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Answer» 1)a1/a2=b1/b2 is not equal to c1/c2then it has infinite number of solution.. 2)a1/a2=b1/b2=C1/c2=1/2then the equation has no solution |
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| 62334. |
\left. \begin{array} { l } { x - 3 y - 3 = 0 } \\ { 3 x - 9 y - 2 = 0 } \\ { 3 x - 5 y = 20 } \\ { 6 x - 10 y = 40 } \end{array} \right. |
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| 62335. |
Express the following minni umbes demais membes1000FIII. . |
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| 62336. |
Anitemmarkedat840 issoldfor Rs 714, What is the discount and discount % |
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Answer» M.P.=840S.P.=714so discount=(840-714)/(840)*100=15% |
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| 62337. |
AICBSE 2015]Find the value of a' so that the angle between the two vectors i + k andi j + a k may be3ition yoctons (111 5 (0, 3, 3),1 5, 9, 3, 3), respectively, then show that |
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| 62338. |
·Find the Simple Interest on6500 at 8% from 5 January 2015 to 19 March 2015 |
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| 62339. |
An item marked at 840 is sold for? 714. what is the discount and discount %? |
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Answer» M.P.=840S.P.=714so discount=(840-714)/(840)*100=15% |
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| 62340. |
A cycle is marked at R 3600 and sold for 3312.What is the discount and discount percentage? |
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| 62341. |
(HCF of 96 and120 is represented as 120x+96y, using Euclid's Division AValue of s |
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| 62342. |
Exercise 8.1breadth.The lensthand breadth of some rectangles are given. Find their perimeter usiof some rectangles are given. Find their perimeter usingthe formula.(a) l=3.5 m, b = 3 m(h) 1- 42 cm h-1 |
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| 62343. |
xample 13 Find r, if 5 4P = 6 5P . |
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Answer» nPr = n!/(n - r)!So, 4Pr = 4!/(4 - r)! and 5P(r - 1) = 5!/(6 - r)! ii) Hence, given equation is: 5*4!/(4 - r)! = 6*5!/(6 - r)!==> 5!/(4 - r)! = 6*5!/{(6 - r)(5 - r)(4 - r)!} ==> 1 = 6/(6 - r)(5 - r)==> 6 = (6 - r)(5 - r)==> 6 = 30 - 11r + r²==> r² - 11r + 24 = 0==> (r - 8)(r - 3) = 0So r = 8 or 3 But r = 8 is not possible, since r has to be less than or equal to 4 Thus r = 3 only |
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| 62344. |
25 x 16 (60 15) -4x (77 62)+ (20 x 6 3) |
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| 62345. |
Is the ratio a rate? Write yes or no and tell why.1. 5 circles to 2 circles2. 365 days per year |
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Answer» Aratiois a comparison of two numbers or measurements. The numbers or measurements being compared are called the terms of the ratio. Arateis a specialratioin which the two terms are in different units. 1. No 2. Yes |
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| 62346. |
word theee lationel umbes between 021 |
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Answer» 0.1, 0.5, 0.8 are 3 rational numbers between 0 &1 To obtain three rationals between 0 and 1 we will multiply by 4/4. 0 × 4/4 = 0/4 1 × 4/4 = 4/4 Three rationals between 0 and 1 are - 1/4 , 2/4 and 3/4. |
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| 62347. |
८८४-१९८३-८।6°4 ,021-Male. |
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| 62348. |
A jacket wasda discountsold for Rs 680. After allowingof 15. an MP. Find MP |
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Answer» Let the Marked price be be a Discount = 15 % of a = 15a/100 Cost price = Marked price - Discount => 680 = a - 3a/20 => 680 = 17a/20 => a = 680 × 20 / 17 => a = 40 × 20 => a = 800 Marked price = Rs 800 Marked price ₹800Cost price =marked price - discount |
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| 62349. |
4 jacket wasva discountsold for Rs 680. After allowinga 15-)) on MP. Field MP 0 |
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Answer» Let the Marked price be be a Discount = 15 % of a = 15a/100 Cost price = Marked price - Discount => 680 = a - 3a/20 => 680 = 17a/20 => a = 680 × 20 / 17 => a = 40 × 20 => a = 800 Marked price = Rs 800 |
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| 62350. |
5.12EXERCISE 25.1given belowy a school boy on different activities in a working day, iActivities / Sleep | School | Home-I-Play-t7Others Total24Number of hours3284Prosont |
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Answer» thank u |
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