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Nostrana [21]
4 years ago
6

Is -5 greater than -9

Mathematics
2 answers:
sveta [45]4 years ago
4 0

Answer: Yes! -5 is greater than -9, because with negatives the smaller numbers are the greatest because they are below 0. Have a good day and Good Luck!


Step-by-step explanation:


Usimov [2.4K]4 years ago
4 0

Answer:

Yes.

Step-by-step explanation:

The reason why is because your going backwards so. If you think about it 5 is less than 9 but -5 is highest than -9

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Answer:

2, 12

Step-by-step explanation:

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LCM of 6 and 8 is 24

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5 0
4 years ago
( 9b - 3 ) ( b + 6 ) <br> Simplify this expression ! <br> Please answer this !!
Colt1911 [192]

Answer:

9b^2+3b-18

Step-by-step explanation:

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7 0
3 years ago
there are 16 boys and 24 girls in chorus whats the percent of the total number of students in chorus are boys?
Ipatiy [6.2K]
The percent of boys in the class is 40% because 16 + 24 = 40 (the entire amount of kids) 16 / 40 = .4 then move decimal twice to the right for percentage which gives you 40%

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I hope this helped :)

5 0
3 years ago
Read 2 more answers
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Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
3 years ago
A total of 50,000 contestants are chosen for a survey. A
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I am pretty sure it is linear decrease,
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