Compare rational numbers what is bigger 4.9 vs 4.6

Mathematics

1 answer:

slamgirl [31]3 years ago

7 0

4.9 is the larger number. Another way to write this is: $4.9 \ \textgreater \ 4.6$

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What is the geometric mean between 1 and 16?

vladimir2022 [97]

-4 is also a geometric mean

if we consider a(first element in series)=1 and r(common ratio)=-4 then the series would be

1,-4,16,-64,256,…..,a(-4)^(n-1 )

where n is nth term

from this -4 would be the geometric mean if we consider -4 as common ratio .

If we consider 4 as common ratio then geometric mean should be 4

so you should mention whether common ratio >0 or not (r>0 or not) .

[without ‘r’ value you can’t solve the question but in general most of the teachers will consider r>0.]

So, -4 won’t be geometric mean of 1 & 16

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3 years ago

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2 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :