The polynomial's maximum value is relative.
<h3></h3><h3>Is the maximum relative or absolute?</h3>Here we have the polynomial:
Notice that the leading coefficient is negative, so, as x tends to negative infinite, V(x) will tend to positive infinite.
So we only have a relative maximum, because is local maximum (at x = 6.4), but the function can take larger values than that.
You can also check that on the graph of V(x), which you can see below:
If you want to learn more about maximums:
#SPJ1
Answer:
Step-by-step explanation:
<h3>1.</h3>The given expression expression can be simplified to ...
3log(10) -2log(5) = log(10^3) -log(5^2) = log(1000) -log(25)
= log(1000/25) = log(40) . . . . ≠ log(5)
≈ 1.60206
Or, it can be evaluated directly:
= 3(1) -2(0.69897) = 3 -1.39794
= 1.60206
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<h3>2.</h3>The properties of logarithms apply to logarithms of any base. Natural logs and common logs are related by the change of base formula ...
ln(x) = log(x)/log(e) ≈ 2.302585·log(x)
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<h3>3.</h3>The given "property" is nonsense. There is no simplification for the product of logs of the same base. There is no expansion for the log of a sum. The formula for the log of a power does apply:
Numerical evaluation of Mr. Kim's expression would prove him wrong.
log(3)log(4) = (0.47712)(0.60206) = 0.28726
log(7) = 0.84510
0.28726 ≠ 0.84510
log(3)log(4) ≠ log(7)
Answer:
Step-by-step explanation:
The standard equation for circle is
where point (a,b) is coordinate of center of circle and r is the radius.
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Given
center of circle =((-2,3)
let r be the radius of circle
Plugging in this value of center in standard equation for circle given above we have
Given that point (1,2 ) passes through circle. Hence this point will satisfy the above equation of circle.
Plugging in the point (1,2 ) in equation 1 we have
now we have value of r^2 = 10, substituting this in equation 1 we have
Thus complete equation of circle is