Multiplying these together, we get:
The square root of x^2 is x, so we have a final expression of 4x.
One way to solve the system is to <u>substitute</u> a variable.
<u>Explanation:</u>
One approach to solve an equation is by substitution of one variable. Right now, a condition for one factor, at that point substitute that arrangement in the other condition, and explain. All value(s) of the variable(s) that fulfills a condition, disparity, arrangement of conditions, or arrangement of imbalances.
The technique for tackling "by substitution" works by settling one of the conditions (you pick which one) for one of the factors (you pick which one), and afterward stopping this go into the other condition, "subbing" for the picked variable and fathoming for the other. At that point you back-explain for the principal variable.
Answer:
m = 6.57 and n = -2
Step-by-step explanation:
To answer this question you will covert the thickness of a dollar to standard notation and then compare the two values.
0.07 =compared to 0.0043 = 0.0657
0.0657 = 6.57 x 10^-2
A quarter is thicker than a dollar by 6.57 x 10^-2.
m = 6.57 and n = -2
Answer:
a) 23.76%
b) 7.8%
Step-by-step explanation:
a) probability that a failure is due to loose keys.
loose key failure (27%) comes under mechanical failure(88%)
hence, probability that a failure is due to loose keys= 0.27×0.88= 0.2376= 23.76%
b) probability that a failure is due to improperly connected wire which comes under electrical failure = 0.12×0.13
probability that a failure is due to poorly welded wires which comes under electrical failure= 0.52×0.12
now, the probability that a failure is due to improperly connected or poorly welded wires. = 0.12(0.52+0.13)= 0.078= 7.8%
So the given series is "16, 06, 68, 88, __"
Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).
From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.
Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.
