The HCF of 4 and 9 is 1. To calculate the Highest common factor of 4 and 9, we need to factor each number (factors of 4 = 1, 2, 4; factors of 9 = 1, 3, 9) and choose the highest factor that exactly divides both 4 and 9, i.e., 1. How to Find the HCF of 4 and 9...
highest common factor hcf or highest common factor is the greatest number which divides each of the two or more numbers. hcf is also called the greatest common measure (gcm) and greatest common divisor(gcd) . hcm and lcm are two different methods, where lcm or least common multiple is ...
HCF of 40, 42 and 45 is the largest possible number which divides 40, 42 and 45 without leaving any remainder. The methods to compute the HCF of 40, 42, 45 are explained here.
Express {(x,y):x2+y2=25wherex,y∈W} as a set of ordered pairs. View Solution Express each of the following as a fractions in the formxywhere x, y ,∈1andy≠0, 2.−476 View Solution Draw the graph of[y]=sinx,x∈[0,2π]where[⋅]denotes the greatest integer function ...
【题目】23) Find the LCM of 9 and 12.24) Find the HCF of 16 and 28.25) Write the product in the boxes. 相关知识点: 试题来源: 解析 【解析】 Icmof 9,12 HCFof 16.2 319.121 4116.281 3.4 4.7 ∴2cm=3*3*4 -HCF 141 =136 反馈 收藏 ...
If (x−k) is the HCF of 3x2+14x+16 and (6x3+11x2−4x−4), then the value of k A23 B2 C−2 D−12Submit Question 2 - Select One If x=43 is a root of the polynomial f(x)=6x3−11x2+kx−20, find the value of k Ak=19 Bk=29 Ck=9 Dk=49Submit Question ...
The In hcf can exchange reversibly with all of these cations. This feature makes the In hcf a good candidate for many electrochemical applications where a source or sink of metal cations is needed, e.g., electrochromic and energy storage devices. A prototype electrochromic system based on an ...
Learn Properties of HCF and LCM and the relation between LCM and HCF of natural numbers with examples. Formula to find HCF and LCM of fractions at BYJU'S.
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To find the value of k given that (x−k) is the highest common factor (HCF) of the polynomials x2+x−12 and 2x2−kx−9, we can follow these steps: Step 1: Set x−k=0Since (x−k) is a factor of both polynomials, we can set x=k. Step 2: Substitute x=k into ...