To round a number to the nearest hundred, we count two places to the left of the decimal point, or from the last digit if the number is a whole number.
If the second digit from the last digit is upto 5, we add 1 to the preceding digit and we complete the last two numbers with zeros.
Therefore, any number from 11,950 to 12,049, will result to 12,000 when rounded to the nearest hundred.
Given data, cos(A - B) = cosAcosB + sinAsinB
<span>let, A=60' and B=30' ( here the ' sign bears degree) </span>
<span>L.H.S = cos(A - B) </span>
<span>=cos (60'-30) ( using value of A and B ) </span>
<span>=cos30' </span>
<span>= sqrt3/2 </span>
<span>R.H.S= cosAcosB + sinAsinB </span>
<span>=cos60' cos30' + sin60' sin30' </span>
<span>= 1/2* sqrt3/2+ sqrt3/2*1/2 </span>
<span>= sqrt3/4 + sqrt3/4 </span>
<span>=2 sqrt3/4 </span>
<span>= sqrt3/2 </span>
<span>so L.H.S =R.H.S or cos(A - B) = cosAcosB + sinAsinB</span>
Elsa's answer is incorrect since there is a solution of the given equation. In the given logarithmic problem, we need to simplify the problem by transposing log2(3x+5) in the opposite side. The equation will now be log2x-log2(3x+5)=4. Using properties of logarithm, we further simplify the problem into a new form log (2x/6x+10)=4. Then transform the equation into base form 10^4=(2x/6x+10) and proceed in solving for x value which is equal to 1.667.
We know that
<span>A number x, rounded to 1 decimal place is 12.3
</span><span>so
x>=12.25
and
x < 12.35
</span><span>the error interval for x is the interval [12.25,12.35)
</span>
the answer is
[12.25,12.35)
