Explore topic-wise InterviewSolutions in Current Affairs.

993x 93x 293x 2 +ax−2=093x 2 +ax−2=0Since, ONE root is 1, then x=193x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−193x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−B=0 has equal roots.93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ SUBSTITUTING a=−1 ]93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=3693x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=36∴ b=9

29 will be the answer because by ADDING 2 and 9 we will GET 11, which will DIVIDE by 11 and the REMAINDER will be 0.

VCCXL IS INVALIDPLEASE MARK ME BRAINLIEST

n other words, the distance surrounding a circle is KNOWN as the CIRCUMFERENCE of the circle. The diameter is the distance across a circle through the centre, and it touches the two POINTS of the circle perimeter.Step-by-step EXPLANATION:

993x 93x 293x 2 +ax−2=093x 2 +ax−2=0Since, one ROOT is 1, then x=193x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−193x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is GIVEN that ax 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6A) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36A 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=3693x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=36∴ b=9

It's simple friend because the denominator is same. -4/12 < -3/12Fact :- In negative INTEGER the smallest negative number is GREATEST or largest negative number. I hope it's help you please MARK me as Brainlist.I need ONE brainlist please :)

In 1 multiplicationmultiplication

ation is the RATE of change of velocity. Usually, acceleration MEANS the SPEED is changing, but not always. When an object moves in a circular path at a constant speed, it is STILL accelerating, because the direction of its velocity is changing.#нσρє íт нєℓρѕ♡~#мαяк αѕ вяαíηℓíѕт☘️plz don't REPORT *innocent faceu*

questions 6 answer OPTION Aquestion 7 answer option CStep-by-step EXPLANATION:HOPE it helps you plz follow and drop me many thanks

rning FRIEND..Your ANSWER is:-12x-9step by STEP explaination:-see attachment So, 12x-9 must be SUBTRACTED from 4x⁴-2x³-6x²+x-5Hope it's helpful for you have a NICE day

7 + 9 + 11 + 11 + 14 + 16 + 17 + 18 + 20 +24 =147Hope it will HELP you

pls mark me as a brainliest friend pls Step-by-step EXPLANATION: FIRST we have to convert the mixed fraction into improper fraction . So , = Therefore the MULTIPLICATIVE INVERSE is .Hope it’s USEFUL friend…Good luck and have a great day buddy…Keep rocking !!

purple{ \Large :\longmapsto \UNDERLINE {\boxed{{\BF {t}^{2} \propto {r}^{3} } }}}:⟼ t 2 ∝r 3 \red{ \underline{ \underline{\LARGE\bf \bigstar \: Hence \: PROVED \: \bigstar}}} ★HenceProved★ -----------------------\large \bf \clubs \: Notations \: Used:-♣ NotationsUsed:−t = Time PERIOD of Revolutionm = MASS of PlanetM = mass of Sunr = radius of Orbitω = Angular FrequencyG = Gravitational Constant

is the FULL FORM of Pagoler UTSAV BEDE GECHE.

* 161 = 93,058b. 5281 * 3491 = 18435971c. 1291 X 592= 764272d.9250 X 29 = 268250Please MARK me As BRAINLIEST

2a²- 2AB Step-by-step explanation:3a² + AB - ( a² + 3AB )= 3a² + ab - a² - 3ab = 2a² - 2ab

-STEP EXPLANATION:Iltutmish organised the Iqta system. He GRANTED LAND to his army. These granted lands were KNOWN as Iqta. The iqta holders were known as Iqtadar or Muqti.

24Step-by-step explanation:16/3 X 9/2= 8 x 3 = 24

I THINK that the NUMBER will CONTAIN a MINUS signeg: 9-²= 9×-9=-81

a)2Y B)2x+3y=280Step-by-step EXPLANATION:

this QUESTION is very IMPORTANT

5.3Step-by-step EXPLANATION:= [ (-48)÷3)÷3​]= [(-16)÷3]= [16÷3]= 5.3

from 1 RUPEES 2 apples can bwe buyStep-by-step EXPLANATION:cost of 20 apples = 10rupeesso, cost of 1 APPLE =20/10= ₹21 apple cost is 2 rupeesplease mark brain list

Name of CountriesBorrowed Features of the Constitution Britain 1. Parliamentary government 2. Rule of Law 3. Legislative procedure 4. Single citizenship 5. Cabinet system 6. Prerogative writs 7. Parliamentary PRIVILEGES 8. Bicameralism Ireland 1. Directive Principles of State POLICY 2. Method of Election of the president 3. Members nomination to the Rajya Sabha by the President Unites States of America 1. Impeachment of the president 2. Functions of president and vice-president 3. Removal of Supreme Court and High court judges 4. Fundamental Rights 5. Judicial review 6. Independence of judiciary 7. The preamble of the constitution Canada 1. Centrifugal form of federalism where the centre is stronger than the states. 2. RESIDUARY powers vest with the centre 3. Centre appoints the Governors at the states 4. Advisory jurisdiction of the supreme court AUSTRALIA 1. Concept of Concurrent list 2. Article 108 i.e. Joint sitting of the two houses 3. Freedom of trade and commerce USSR (Now Russia) 1. Fundamental duties 2. The ideals of justice (social, economic and political), expressed in the Preamble. France 1. Concept of “Republic” 2. Ideals of Liberty, Equality and Fraternity(contained in the Preamble) Germany 1. Fundamental Rights are suspended during Emergency South Africa 1. Election of members of the Rajya Sabha 2. Amendment of the Constitution Japan 1. Concept of “procedure established by Law”

option 3 is correct see FIRST 1 zero than -1 = O than second -1= O Step-by-step explanation:PLEASE mark me as BRAINLIST please

are root of 3480 is a RATIONAL NUMBER if 3480 is a perfect square. It is an irrational number if it is not a perfect square. Since 3480 is not a perfect square, it is an irrational number. This MEANS that the answer to "the square root of 3480?" will have an infinite number of DECIMALS. The decimals will not terminate and you cannot make it into an exact fraction.

(a-3)³Step-by-step EXPLANATION:the ANSWER is (a-3)³ as the EQUATION COMES down to that PROPERTY

2x+3+\frac{X}{5}=7+2xx+3+5x=7+2xSolve1Find COMMON denominator+3+5=7+2x+3+\frac{x}{5}=7+2xx+3+5x=7+2x55+5⋅35+5=7+2

Legumes form a unique symbiotic relationship with bacteria known as RHIZOBIA, which they allow to INFECT their roots. This leads to root nodule formation where bacteria are accommodated to CONVERT NITROGEN from the air into ammonia that the plant can use for growth.

the scientific knowledge and/or EQUIPMENT that is needed for a particular INDUSTRY, etc.विशिष्‍ट उद्योग आदि के लिए अपेक्षित वैज्ञानिक जानकारी और/या उपकरण; प्रौद्योगिकी/शिल्‍पविज्ञान और/या तकनीकी उपकरण

:−\BEGIN{gathered} \sf{2A + 2B = \left[\begin{array}{cc}2& - 1& 4 \\3 & 2 & 5\end{array}\right] } \\ \\ \sf{A + 2B = \left[\begin{array}{cc}5 & 0 & 3 \\1 & 6 & 2 \end{array}\right]}\end{gathered} 2A+2B=[ 23 −12 45 ] A+2B=[ 51 06 32 ] -----------------------\large \bf \clubs \: To \: Find :-♣ ToFind:−Value of 2B .-----------------------\large \bf \clubs \: Solution:-♣ Solution:−We Have,\begin{gathered} \pmb{2A + 2B = \left[\begin{array}{cc}2& - 1& 4 \\3 & 2 & 5\end{array}\right] } \: \: - - - - (1)\\ \\ \: \: \: \: \:\pmb{A + 2B = \left[\begin{array}{cc}5 & 0 & 3 \\1 & 6 & 2 \end{array}\right]} \: \: - - - - (2)\end{gathered} 2A+2B=[ 23 −12 45 ]2A+2B=[ 23 −12 45 ] − − − −(1) A+2B=[ 51 06 32 ]A+2B=[ 51 06 32 ] − − − −(2) Multiplying Eq. (2) by 2 We Get :\begin{gathered} \pmb{2A + 4B = \left[\begin{array}{cc}10 & 0 & 6 \\2 & 12 & 4\end{array}\right]} \: \: - - - - (3)\end{gathered} 2A+4B=[ 102 012 64 ]2A+4B=[ 102 012 64 ] − − − −(3) Subtracting (1) From (3) We Get :\begin{gathered} \sf \cancel{2A} + 4B - \cancel{2A} -2B \\ = \left[\begin{array}{cc}10 & 0 & 6 \\ 2 & 12 & 4\end{array}\right] - \left[\begin{array}{cc}2 & - 1 & 4 \\3 & 2 & 5\end{array}\right] \\ \\ :\longmapsto\sf2B = \left[\begin{array}{cc} 10- 2& 0+ 1& 6- 4 \\ 2- 3 & 12- 2 & 4 - 5\end{array}\right] \\ \\ \end{gathered} 2A +4B− 2A −2B =[ 102 012 64 ]−[ 23 −12 45 ] :⟼2B=[ 10−22−3 0+1 12−2 6− 44−5 ] \begin{gathered}\purple{ \large :\longmapsto \pmb{ \underline {\BOXED{{2B = \left[\begin{array}{cc} 8& 1& 2 \\ - 1 & 10 & - 1\end{array}\right]} }}}}\end{gathered} :⟼ 2B=[ 8−1 110 2 −1 ] 2B=[ 8−1 110 2 −1 ] \underline{\underline{\Large\pink{\mathfrak{ \text{H}ence \:\:option\:\: B\:\: Correct} }}} HenceoptionBCorrect \begin{gathered} \Large\red{\mathfrak{ \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}\end{gathered} Whichistherequired Answer. -----------------------

(X^n=0) x= all REAL number1+39+6+79 40+85 125