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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 22201. |
The distance between the point (X, 2) and (3,-6) is 10 unit what is the postive value of x |
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Answer» Hii Beauty parlor facial ka rate kya hai X=9 Distance formula lagao equal me 10 rakh ke solve karlo |
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| 22202. |
Frequently asked questions in exams |
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Answer» Aur kya Hi ???guys Hi Mr. Sahil There are many questions.....yha lamba ho jayega likhna.. |
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| 22203. |
Find the value of k for which one root of the quadratic equation kx -14+8=0 |
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Answer» Hii kx - 14 + 8 = 0kx - 6 = 0kx = 6k = 6/x I think that your question is wrong❌ Equation is wrong ❌ |
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| 22204. |
If the figure,Am perpendicular to Bc .If tanB=3/4 ,tanC=5/12 and Bc=56cm.find the length of Am |
| Answer» In right\xa0{tex}\\triangle {/tex}AMB,\xa0tan B =\xa0{tex}\\frac { 3 } { 4 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { A M } { B M } = \\frac { 3 } { 4 }{/tex}{tex}\\Rightarrow{/tex}\xa04AM = 3BM {tex}\\Rightarrow{/tex}BM =\xa0{tex}\\frac { 4 } { 3 }{/tex}AM ...(i)In right\xa0{tex}\\triangle{/tex}AMC,tan C =\xa0{tex}\\frac { \\mathrm { AM } } { \\mathrm { MC } }{/tex}{tex}\\Rightarrow \\frac { 5 } { 12 } = \\frac { \\mathrm { AM } } { \\mathrm { MC } }{/tex}{tex}\\Rightarrow{/tex}\xa0MC =\xa0{tex}\\frac { 12 } { 5 }{/tex}\xa0AM ...(ii)Now, BM + MC = BC{tex}\\frac { 4 } { 3 }{/tex}AM +\xa0{tex}\\frac { 12 } { 5 }{/tex}AM = 56AM\xa0{tex}\\left( \\frac { 4 } { 3 } + \\frac { 12 } { 5 } \\right){/tex}\xa0= 56AM\xa0{tex}\\left( \\frac { 20 + 36 } { 15 } \\right){/tex}\xa0= 56{tex}\\Rightarrow{/tex}\xa0AM =\xa0{tex}\\frac { 56 \\times 15 } { 56 }{/tex}= 15 cm | |
| 22205. |
The coordinate of circumcentre of triangle formed by points O(0,0),A(a,0)&B(0,b)are |
| Answer» The coordinate of A (a,0), B (0,b) and O (0, 0)Area of Triangle AOB\xa0{tex} = \\frac{1}{2} \\times base \\times height{/tex}{tex} = \\frac{1}{2} \\times a \\times b = \\frac{1}{2}ab\\,sq.\\,units{/tex} | |
| 22206. |
Solve the equation by cross multiplication methodX+Y=a+bax_by=(a+b)(a_b) |
| Answer» X | |
| 22207. |
If cosβ+sinβ=√2 cos β show that cosβ-sin=√2 sinβ |
| Answer» cosB+sinB=√2cosBsinB=√2cosB-cosBsinB=cosB(√2-1). multiply by (√2+1)sinB×(√2+1)=cosB(√2-1)×(√2+1)√2sinB+sinB=cos(2-1)√2sinB=cosB-sinBHo gya ?? | |
| 22208. |
Can anyone suggest some important trigonometry proof questions? |
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| 22209. |
In board exam only ncert questions will come or not |
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Answer» Not completely from ncert, But 10 previous year and ncert is enough? yes |
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| 22210. |
In a triangle right angled at B.if Ab is 3 cm and Bc is 6 ,then find angle a and b |
| Answer» Gadit lag rahi hai bhaiya | |
| 22211. |
Out of course 4444=20 using addition,subtract,multiple and division |
| Answer» [(4÷4)+4]×4 = 20 | |
| 22212. |
If (x-√5) is factor of the cubic polynomial x³-6x+9 then find the value of k |
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Answer» Hint krdo na Sorry yar typing mistake Please check yur question, there is no k in yur polynomial. Mere ko toh k nhi dikh rha |
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| 22213. |
Which is better rd or rs |
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Answer» I think rs,ncertis better than rd Rd n ncert is better First complete your ncert, after that r.d is better in my point of view. i think RS is better than RD NCERT |
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| 22214. |
The difference of two numbers is 4 and difference of their reciprocal is 4/11find the numbers |
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Answer» Thnx hima ya honey is ryt.. Take x-y=4and for reciprocal 1/y-1/x=4/11 |
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| 22215. |
Root 1 + cos theta by 1 minus cos theta equal cosec theta + cot theta |
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| 22216. |
If edges of cube is increase by 10 percent what percentage LSA is incereased. |
| Answer» Let ‘a’ be the edge of the cube.Surface area of cube = 6a2 sq unitsGiven length of edge is increased by 10%Hence new measure of edge = a + 10% of a = (11a/10)Surface area of cube (after increase) = 6(11a/10)2= (121/100) x 6a2 sq unitsChange in surface area = [(121/100) x 6a2 ] – 6a2 = (21/100) x 6a2% increase = [(21/100) x 6a2] x 100/ [6a2 ]\xa0= 21% | |
| 22217. |
Find the hcf and lcm of 8/9 3/12 and2/3 |
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| 22218. |
What is the difference between mount and surmount |
| Answer» Mounted on (something) - it means that the object is placed over another object.Surmounted by (something) - it means that it is placed over something. | |
| 22219. |
Two horses |
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Answer» Ye question hi nhi h to samajh me kya aayega? Where is two horses? Kya ? |
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| 22220. |
Tan45+sin45=1 |
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Answer» to kk.... ky huaaaa gussa ho gye puja ji y m hk ye misprint h ky I think there is a fault in the question Why dont you check it out |
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| 22221. |
3n^2+5n=sn |
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| 22222. |
How many three digits number divisible by 7 |
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Answer» Sorry that is 128 118 128 |
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| 22223. |
What is matrics |
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| 22224. |
How to find angle of elevation and depression |
| Answer» The angle between line of sight and elevating/depression line | |
| 22225. |
how to condtruct a circle |
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Answer» Naam suna hai... Shayad sanjay dutt ki hai.. May be Ji ab raaj kaun karega ye mai kaise bata sakti Hu... Ye to bhagwan decide karenge Tu kar lena bhai. Kabir is also ryt Aaj english ka preboard tha mera Take length of radius with help of ruler put the point of compass aur ghuma dena ban jayega |
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| 22226. |
Who is the father of mathematics in India... |
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Answer» Google pe aap check kr lijiye... Mai to confident Hu ki mera answer sahi hai According to chemistry, it is the ratio of the atom present in a compound and according to maths it is the relationship between mean mode and median Ratio of atoms present in a compound The formula which states the relationship between mean median and mode. Aryabhatt |
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| 22227. |
What is formula for circumference of semi ciecle? |
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Answer» Circumference of semi circle = 2\xa0× πrCircumference of semi circle = 1/2 × 2\xa0πr = πr Pi.r+diameter |
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| 22228. |
Find the value of k for which one root of the quadratic equation kx-14x+8=0 is 2 |
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Answer» K=10 kx - 14x + 8 = 0If x = 2k(2) - 14(2) + 8 = 02k - 28 + 8 = 02k - 20 = 02k = 20k = 10\xa0 Sorry, it will be K=10. My previous answer was wrong. F(x)=kx-14x+8=0 =K(2)-14(2)+8=0 =2k-28+8=0 =2k-36=0. =2k=36 =K=36÷2 K=18 Let P(x)=kx-14x+8=0=P(2)=k(2)-14(2)+8=0=2k-28+8=0=2k-20=0=2k=20=K=10 K=18. |
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| 22229. |
Lateral surface area of cuboid... |
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Answer» 2(l+b)h 2×height(length +breadth) Your welcome.. Here l = length ; b = breadth ; h = height 2l(b+h) |
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| 22230. |
Show that the points (1,7),(4,2),and (-4,4) are the vertices of squre |
| Answer» There are only three coordinates how can we prove it | |
| 22231. |
If the equation (1+m^2)x^2+2mcx+c^2-a^2=0 has coincident roots show that c^2=a^2(1+m^2) |
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Answer» Laga Di Maine aapni dp puja ji ? Par aap ek msg toh Karo Hamm bhi aapko dekhna chahate hai Puja ji ? Hlo guys can i join this Aacha thik hai Puja ji ?? Pic dekhni hai aapko koi nhi aap ek msg Karo Mai aapko pic send Kar dunga puja ji ? Hlo guys Yaha aajao aap Sab ????? |
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| 22232. |
What is the value of π |
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Answer» 22/7 Mar jao jake 22/7....(3.14 if its given in ques)??.. 22/7 |
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| 22233. |
Mode niklna |
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Answer» The formula is l+[f1-f0/2f1-f0-f2]*h Using by this formula.....l+[f1-f0/2f1-f0-f2]×h |
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| 22234. |
Can anyone tell me how to solve trigonometry easily??? |
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Answer» Practice more, and try to convert the given into the proof yu want.............? Just try to convert everything to sin and cos Trigonometry is only based on its formulas U have to learn...all the formula nd table of trigonometry then u can solve easily...? |
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| 22235. |
√tanAtanB+tanAcotB÷sinAsecB - sin2B÷cos2A= tanA and A+B=90° |
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| 22236. |
If (costheta+sintheta)=root2 sintheta. To prove that (sintheta-costheta)=root2 costheta....? |
| Answer» Give me answer fast? | |
| 22237. |
What is a log |
| Answer» Logarithm .....Logarithmic functions are the inverses of exponential functions. The inverse of the exponential function y = ax is x = ay. The logarithmic function y = logax is defined to be equivalent to the exponential equation x = ay. ... It is called the logarithmic function with base a. U will study in class 11 | |
| 22238. |
X suare value if x is 4 |
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Answer» The correct answer is 16 16 16 will be the answer |
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| 22239. |
ABC is an isosceles triangle right angled at C. Prove that AB sq= 2AC sq |
| Answer» AB sq.=AC sq.+BC sq. And AC=BC hain | |
| 22240. |
Q prove that parallelogram circumscribing a circle is a rhombus?? |
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Answer» Since ABCD is a parallelogram,AB = CD ---- i)BC = AD ---- ii)\xa0It can be observed thatDR = DS (Tangents on the circle from point D)CR = CQ (Tangents on the circle from point C)BP = BQ (Tangents on the circle from point B)AP = AS (Tangents on the circle from point A)Adding all these equations, we obtainDR + CR + BP + AP = DS + CQ + BQ + AS(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)CD + AB = AD + BCOn putting the values of equations (1) and (2) in this equation, we obtain2AB = 2BCAB = BC …(3)Comparing equations (1), (2), and (3), we obtainAB = BC = CD = DAHence, ABCD is a rhombus. Kaku bhai i am not satisfied on your answer. Plz solve it in detail with proper explaination. ab + cd = bc + ad== ab + ab = bc + bc.....(ab = cd, bc = ad)== 2ab = 2bc== ab = bc |
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| 22241. |
Area of similar triangle is equal to the square of the ratio of their corresponding sides |
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Answer» Ncert book mein diya toh hain If u want the proof then it\'s too long I can\'t solve it here... Sorry Ratio of area of sim................................ |
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| 22242. |
If alpha and beta are zeroes of x2-x-1,find the value of 1/alpha+1/beta |
| Answer» -1 | |
| 22243. |
√3-2√3+√3÷-2√3 |
| Answer» | |
| 22244. |
If alpha and beta are zeros of polynomial f(x) = ax2+bx+cFind alpha square minus beta square |
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| 22245. |
Find the value of k so that the area of triangle with vertices (k+1,1),(4,-3 )&(7 -k) |
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Answer» 3 Put the formula, area of triangle =1/2[x1(y2 -y3) + x2(y3-y1) + x3(y1-y2)]...... yu will get the answer........ 6 square units |
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| 22246. |
Find the value of k in eqx2+5kx+16=0 where roots are equal |
| Answer» 8/5 | |
| 22247. |
Find the 20th term from the last term of the AP : 3,8,13, . . ., 253 |
| Answer» Given series is 3, 8, 13, ...... 253here, first term , a = 3common difference , d = 5Let us find total number of terms at first.use an = a+(n-1)d}⇒ 253 = 3 + (n - 1)5⇒ 250 = 5(n - 1)⇒ n - 1 = 50⇒ n = 51so, there are 51 terms in given series.now we know, mth term from last =last term - mth term + 1so, 20th term from last = 51 - 20 + 1 = 32hence, 20th from last = 32th term from firstuse , a32=a+(32-1)d}= 3 + 31 × 5= 3 + 155= 158hence , 20th term from last = 158 | |
| 22248. |
Derivation of volume of frustum |
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| 22249. |
The line segment AB joining A(2,-4)and B(5,2)is trisected at C(3,a) and D(b,0). Find a and b. |
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Answer» I mean uttar Pradesh By formula Teri |
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| 22250. |
Ch 12 ki ex 12.3 ka q 8 |
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Answer» It is already provided in this app ...... Plz check once Mycbseguide me solution h jakar check karo |
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