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| 17301. |
C.p. =425 profit =12•/• find the selling price if |
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Answer» 476 Let the selling price (S. P.) be x.\xa0Hence, selling price is ₹ 476. |
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| 17302. |
S12=3(S8-S4) |
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| 17303. |
PNHG-TRRA-4XR2-HB66D |
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| 17304. |
Given 15 cotA=8, find sinA and secA |
| Answer» Given, 15cotA=8CotA=8/15CotA=AB/BCAB=8,BC=15In triangle ABC,By Pythagoras theorem,AC^2=AB^2+BC^2=(8)^2+(15)^2=64+225=289AC=√289AC=17 unitsSinA=BC/AC =15/17SecA=AC/AB =17/8 | |
| 17305. |
Draw and colour to find the height and distances using trigonometric ratios without actual measuring |
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| 17306. |
In the figure AB||CD tyen the value of X is |
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Answer» The X value is 10 The X value is 10 |
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| 17307. |
Find the zeroes of 6x^2+7x+2 and verify the relationship and it\'s coefficients |
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| 17308. |
R S aggarwal arithmeticprogression all solve questions |
| Answer» Lusu | |
| 17309. |
Natue of root formula |
| Answer» The nature of the roots depends on the value of b2\xa0– 4ac. bx2\xa0– 4ac is called the\xa0discriminant\xa0of the quadratic equation ax2\xa0+ bx + c = 0 and is generally, denoted by D.∴\xa0D = b2\xa0– 4acIf D > 0, i..e., b2\xa0– 4ac > 0, i.e., b2 – 4ac is positive; the roots are\xa0real and unequal. Also,(i) If b2\xa0– 4ac is a perfect square, the roots are\xa0rational and unequal.(ii) If b2\xa0– 4ac is positive but not perfect square, the roots are\xa0irrational and unequal.If D = 0, i.e., b2\xa0– 4ac = 0; the roots are\xa0real and equal.If D < 0, i.e., b2 – 4ac < 0; i.e., b2\xa0– 4ac\xa0is negative; the roots are not real, i.e., the roots are\xa0imaginary. | |
| 17310. |
Nature of roots |
| Answer» Nature of RootsA quadratic equation ax2\xa0+ bx + c = 0 has(i) two distinct real roots, if b2\xa0– 4ac > 0,(ii) two equal real roots, if b2\xa0– 4ac = 0,(iii) no real roots, if b2\xa0– 4ac < 0.Since\xa0b2\xa0– 4ac\xa0determines whether the quadratic equation\xa0ax2 +\xa0bx\xa0+\xa0c\xa0= 0 has real roots or not,\xa0b2\xa0– 4ac\xa0is called the\xa0discriminant\xa0of this quadratic equation | |
| 17311. |
Class 10 exercise 8.4 ques 5 |
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Answer» (i) (cosec θ – cot θ)2\xa0= (1-cos θ)/(1+cos θ)To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)L.H.S. = (cosec θ – cot θ)2The above equation is in the form of (a-b)2, and expand itSince (a-b)2\xa0= a2\xa0+ b2\xa0– 2abHere a = cosec θ and b = cot θ= (cosec2θ +\xa0cot2θ – 2cosec θ cot θ)Now, apply the corresponding inverse functions and equivalent ratios to simplify= (1/sin2θ\xa0+ cos2θ/sin2θ – 2cos θ/sin2θ)= (1 + cos2θ – 2cos θ)/(1 – cos2θ)= (1-cos θ)2/(1 – cosθ)(1+cos θ)= (1-cos θ)/(1+cos θ) = R.H.S.Therefore, (cosec θ – cot θ)2\xa0= (1-cos θ)/(1+cos θ)Hence proved.(ii) (cos A/(1+sin A))\xa0+ ((1+sin A)/cos A) = 2 sec ANow, take the L.H.S of the given equation.L.H.S. = (cos A/(1+sin A))\xa0+ ((1+sin A)/cos A) = [cos2A\xa0+ (1+sin A)2]/(1+sin A)cos A = (cos2A + sin2A + 1\xa0+ 2sin A)/(1+sin A) cos ASince cos2A + sin2A = 1, we can write it as = (1\xa0+ 1 + 2sin A)/(1+sin A) cos A = (2+ 2sin A)/(1+sin A)cos A = 2(1+sin A)/(1+sin A)cos A = 2/cos A = 2 sec A = R.H.S.L.H.S. = R.H.S.(cos A/(1+sin A))\xa0+ ((1+sin A)/cos A) = 2 sec AHence proved.\xa0For more click on the given link:NCERT Solutions for Class 10 Maths Exercise 8.4 ... Yes We have to solve all parts of question 5 |
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| 17312. |
The length of a chain used as boundary of the semi circular park is 90cm .Find the area of the park |
| Answer» Length of the boundary of the semicircular park = 90 m⇒\xa0πr + 2r = 90⇒ r(22/7 + 2) = 90⇒ r = (90*7)/36⇒ r = 17.5 mArea of the semicircular park = 1/2πr²= 1/2*22/7*17.5*17.5= 481.25 sq m | |
| 17313. |
Solve for x 1÷2x-3+1÷x-5=10÷9 |
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Answer» 1/2x - 3+1/x -5=10/91/2x -3+1/x =10/9+51/2x -4/x=50/91/x -1/x=50/9*2/42(1/x) =50/181/x =25/18X= 18/25 ¿ |
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| 17314. |
5555555+66666 |
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Answer» 56,22,221 562,221 5\xa05 5 5 5 5 5 + 6 6\xa066 6 =5 6 2 2 2 2 1 |
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| 17315. |
6/x+y = 7/x-y + 31/2(x+y) = 1/3(x-y)Find the value of (x,y) by elimination method |
| Answer» Hence, solution of the system of equation is x= -5/4, y = -1/4\xa0 | |
| 17316. |
If the equation x2-bx+1=0 has two same roots then |
| Answer» Plz give me ans | |
| 17317. |
Tangent is what? |
| Answer» A line,which touches a circle at a point and that point is perpendicular to the radius of the circle. | |
| 17318. |
root 21 |
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Answer» 4 . 58 Root 21 = root 3×7 |
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| 17319. |
3ROOT3 |
| Answer» 27 | |
| 17320. |
2.040040004 |
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| 17321. |
rational or irrational q no 7 1.535335333 |
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| 17322. |
if a point (x,y) is equidistant from the point Q(9,8) and S(17,8) . Then write the value of (x,y). |
| Answer» Hiii | |
| 17323. |
The pair of equation y=0 and y= -7 has |
| Answer» No solution | |
| 17324. |
Given An=4, d=2, Sn= -14 . Find n and a |
| Answer» N=7, A=-8 | |
| 17325. |
When the term of the AP:3,8,13,18,......,is 78? |
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Answer» 16th term in given A.P is 78Step-by-step explanation:Given\xa0A.P\xa0:\xa03,8,13,18,....,78First\xa0term\xa0(a)\xa0=\xa03\xa0,common\xa0difference\xa0(d)=\xa0= $8-3$= $5$Therefore,16th\xa0term\xa0in\xa0given\xa0A.P\xa0is\xa078 16th term |
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| 17326. |
HCF of 26 and 91 |
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Answer» 13 HCF is 13 Answer :- The HCF of 26 : 2✖13The HCF of 91 : 7 ✖ 13 So 13 is common that\'s why 13 is the HCF. 26 = 2 x 1391 =7 x 13HCF = 13 HCF of 26 and 91 |
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| 17327. |
What is the probability of getting two heads when two coins are tossed together? |
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Answer» 1/4 When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where\xa0H\xa0is denoted for head and\xa0T\xa0is denoted for tail. |
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| 17328. |
. For what value of X, the mode of the following data is 8? 4,5,6,8,5,4,8,5,6,X,8 |
| Answer» 4 | |
| 17329. |
Find the volume of the largest sphere which is carved out of cube of side 7 Cm |
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Answer» R (radius) = 7/2= 3.5Volume of cube = 4/3πr^34/3 ✖ 22/7✖ 3.5 ✖ 3.5 ✖ 3.5Answer = 179.67 cm^3 Edge of the cube = diameter of sphere = 7 cmRadius of the sphere =\xa0\xa0= 3.5 cmVolume of the sphere =\xa0= 179.67\xa0 |
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| 17330. |
If the sum of the first m term of the A.P is 2m+3 , then what is it second term? B |
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| 17331. |
Solve the polynomial 2x2-8x+6 |
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Answer» 10-8x _4x + 6 -5/4 Question 4 |
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| 17332. |
MANANNNXX.13 IS MY ID IN INSTAGRAM DM ME IF U WANT THE HACKED VERSION OF AMONG US |
| Answer» \tDon\'t post personal information, mobile numbers and other details.\tDon\'t use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.\tBe nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.\tAsk specific question which are clear and concise.If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html | |
| 17333. |
a quadratic polynomial can have at most……..zeroes. |
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Answer» 2 zeros 2 |
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| 17334. |
Has the rational number aterminatingoranonterminatingdecimalrepresentation.Givereason |
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| 17335. |
jk |
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| 17336. |
ko |
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| 17337. |
if mean of the following data is 14.7and total frequency is 40.find the value p and q |
| Answer» class freq total0-6 10 3 * 106-12 p 9 * p12-18 4 4 * 1518-24 7 7 * 2124-30 q 27 * q30-36 4 4 * 3336-42 1 1 * 39=====================total 40 408 + 9p + 27 q============================total of frequencies = 26 + p + q = 40 => p+q = 14sum = 408 + 9 p + 27 q = 14.7 * 40 = 588So finally, 9 p + 27 q = 180 => p + 3 q = 20 => 14- q + 3 q = 20 =>\xa0\xa0q = 3 and p = 11 . | |
| 17338. |
Which term of the AP : 3,15,27,39,... will be 132 more than its 54th term? |
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Answer» 65th term Hi a = 3, d = 12 first find 54th term so n = 54 . By formula An = a +(n - 1) d so An = 3+(54-1)12 = 639. To find the term from question, 639+132= 771. By using same formula, 771=3+(n-1)12 we get n = 65. |
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| 17339. |
How cutte were formed? |
| Answer» What??? | |
| 17340. |
What is triangle congratulations property |
| Answer» CPCT Rules in MathsThe full form of CPCT is Corresponding parts of Congruent triangles. Congruency can be predicted without actually measuring the sides and angles of a triangle. Different rules of congruency are as follows.\tSSS (Side-Side-Side)\tSAS (Side-Angle-Side)\tASA (Angle-Side-Angle)\tAAS (Angle-Angle-Side)\tRHS (Right angle-Hypotenuse-Side) | |
| 17341. |
Triangles theorem 6.7 |
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Answer» if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangle of the both sides of the perpendicular are similar to the whole triangle and to each other Ans :- if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangle of the both sides of the perpendicular are similar to the whole triangle and to each other |
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| 17342. |
272 |
| Answer» Ka ha ya | |
| 17343. |
A/(ax-1)+ b/(bx-1)=(a+b) |
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| 17344. |
The multiplicative inverse of zeroAdoes not existB1CD0 |
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Answer» Not exist A) does not exist |
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| 17345. |
1+333333 |
| Answer» 333334 | |
| 17346. |
Solve for x: 2 * [x ^ 2 + 1/(x ^ 2)] - 9[x + 1/x] + 14 = 0 |
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| 17347. |
This year sample paper? |
| Answer» Sample paper of this year are available on this app..below the homework help section....? | |
| 17348. |
Koi i n s t a pr h kya |
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| 17349. |
find the value of k for which the quadratic equation 2x^2+kx+2=0 has equal roots |
| Answer» The values of k = 4.Given:Quadratic equation\xa0\xa0has equal roots.To find:The values of kExplanation:General format of quadratic equation is\xa0.If the roots of quadratic equation is equal then, Where, a = coefficient of x² b = coefficient of x c = constant term. b² - 4ac =0Comparing the quadratic equationb = k a=2 c=2 k² - 4×2×2 = 0 k² = 16 k =\xa0k = 4Therefore, the values of k = 4 if roots are equal. | |
| 17350. |
If 4 tan =3,evaluate 4 sin-cos+1 4cos+cos-1 |
| Answer» 4tan = 3 ⇒ tan = 3/4we know, tan = perpendicular/baseso, after comparing , perpendicular = 3 and base = 4∴ hypotenuse = √(3² + 4²) = 5now, sin = perpendicular/hypotenuse = 3/5cos = base/hypotenuse = 4/5now, (4sin - cos + 1)/(4sin + cos - 1)= (4 × 3/5 - 4/5 + 1)/(4 × 3/5 + 4/5 - 1)= (12 - 4 + 5)/(12 + 4 - 5)= 13/11 | |