Answer:
14 is the GCF of 28 and 98
Answer:
Those value satisfy the equation are solution
So, (0,7/5), (7/3,0) , (9,4) are the solution
No, mistake at s2, if there was 0x then they shouldn’t have written anything at all because there are 0 x’s!!!
1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2
Answer:
d1=2A/d2
d2=2A/d1
Step-by-step explanation:
The options of this question are:
1. d₁=2Ad₂
2. d₁= 2A/d₂
3. d₂= d₁/2A
4. d₁= 2A/d₂
5. d₂= 2Ad₁
Given:
A=1/2(d1*d2)
Multiply both sides by 2
We have,
2A=d1*d2
Divide both sides by d2
2A/d2=d1*d2/d2
2A/d2=d1
Therefore, d1=2A/d2
Similarly, from the previous equation
2A=d1*d2
Divide both sides by d1
2A/d1=d1*d2/d1
2A/d1=d2
Therefore,
d2=2A/d1
Options
2. d₁= 2A/d₂
4. d₁= 2A/d₂
