Round 70.74 to 2 significant figures

1 answer:

8 0

The answer is 71 since you count from the tenth space. So the first significant figure would be 7 and the second significant figure would be 0. As we know that we round up one number of the previous number is 5 or bigger.

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Find n 5|6n+15|+2=-43

Ivenika [448]

Answer:

<h2>No solution</h2>

Step-by-step explanation:

$5|6n+15|+2=-43\qquad\text{subtract 2 from both sides}\\\\5|6n+15|+2-2=-43-2\\\\5|6n+15|=-45\qquad\text{divide both sides by 5}\\\\\dfrac{5\!\!\!\!\diagup^1|6n+15|}{5\!\!\!\!\diagup_1}=\dfrac{-45\!\!\!\!\!\diagup^9}{5\!\!\!\!\diagup_1}\\\\|6n+15|=-9\qquad\bold{FALSE}\to\text{no solution}\\\\\text{because}\ |a|\geq0\ \text{for any real value of}\ a.$

3 0

3 years ago

Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at the

scZoUnD [109]

Answer:

$\frac{y}{x+y}$

Step-by-step explanation:

The required answer is the rate at which Machine A works when the two machines are combined.

Note: the rate of doing work is express as

$rate=\frac{1}{time taken} \\$

Hence we can conclude that Machine A working rate is

$machine A=\frac{1}{x} \\$ and machine B working rate is

$machine B=\frac{1}{y} \\$

When the two machine works together, the effective working rate is

$\frac{1}{x}+\frac{1}{y}\\\frac{xy}{x+y}\\$

The fraction of the work that Machine B will not have complete because of Machine A help is the total work done by machine A

Hence the fraction of work done by A is expressed as

$\frac{1}{x}*combine working rate$

$\frac{1}{x}*\frac{xy}{x+y}\\\frac{y}{x+y} \\$

Hence the fraction of the work that Machine B will not have complete because of Machine A help is the total work done by machine A is $\frac{y}{x+y} \\$