Then, | x | - | x | = 2 |
10 | 15 |
⟹ 3x - 2x = 60
⟹ x = 60 km.
Time taken to travel 60 km at 10 km/hr = | ❨ | 60 | ❩hrs | = 6 hrs. |
10 |
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
∴ Required speed = | ❨ | 60 | ❩kmph. | = 12 kmph. |
5 |
We know that In center of every triangle ( acute , obtuse and right angle ) lies in its interior.
Hence required answer will be any triangle .
If (5x2 ? 3y2) : xy = 11 : 2, then the positive value of |
|
is : |
y |
4.2 x 4.2 - 1.9 x 1.9 | is equal to: |
2.3 x 6.1 |
Given Expression = | (a2 - b2) | = | (a2 - b2) | = 1. |
(a + b)(a - b) | (a2 - b2) |
(7 + 3√5)(7 - 3√5) | = | (7)2 - (3√5)2 | = √49 - 45 = √4 = 2. |
If √5 = 2.236, then the value of | √5 | - | 10 | + √125 is equal to: |
2 | √5 |
√5 | - | 10 | + √125 | = | (√5)2 - 20 + 2√5 x 5√5 |
2 | √5 | 2√5 |
= | 5 - 20 + 50 |
2√5 |
= | 35 | x | √5 |
2√5 | √5 |
= | 35√5 |
10 |
= | 7 x 2.236 |
2 |
= 7 x 1.118 |
= 7.826 |
If log10 7 = a, then log10 | ❨ | 1 | ❩ | is equal to: |
70 |
a |
10 |
1 |
10a |
|
= log10 1 - log10 70 | |||||
= - log10 (7 x 10) | ||||||
= - (log10 7 + log10 10) | ||||||
= - (a + 1). |
log √8 | is equal to: |
log 8 |
1 |
√8 |
1 |
4 |
1 |
2 |
1 |
8 |
1 |
2 |
log √8 | = | log (8)1/2 | = | ½log 8 | = | 1 | . |
log 8 | log 8 | log 8 | 2 |
The value of | ❨ | 1 | + | 1 | + | 1 | ❩ | is: |
log3 60 | log4 60 | log5 60 |
Given expression | = log60 3 + log60 4 + log60 5 |
= log60 (3 x 4 x 5) | |
= log60 60 | |
= 1. |
.009 | = .01 |
? |
Let | .009 | = .01; Then x = | .009 | = | .9 | = .9 |
x | .01 | 1 |
Evaluate : | (2.39)2 - (1.61)2 |
2.39 - 1.61 |
Given Expression = | a2 - b2 | = | (a + b)(a - b) | = (a + b) = (2.39 + 1.61) = 4. |
a - b | (a - b) |
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.