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Afina-wow [57]
3 years ago
9

yuna missed Five Points out of 100 points on her math test what decimal number represents the part of her math test that she ans

wered correctly
Mathematics
2 answers:
vfiekz [6]3 years ago
8 0
The number that represents what she got correct is 0.95, because 100-5=95, and because that the number will be 0.95, if she had gotten all correct the number would have been a full 1
Naya [18.7K]3 years ago
8 0

THE answer would be .95 because 100 - 5 = 95 or 95/100 = .95.


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Help please<br> Calculate the product of 8/15, 6/5, and 1/3. 
bogdanovich [222]

Hey there!

Great question:)


8/15 * 6/5 * 1/3 = 16/75


I hope this helps;)

4 0
3 years ago
I think I found the area I need help finding the perimeter
IrinaK [193]
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Perimeter is amount out all sides added together.
8 0
3 years ago
NO LINKS OR FILES!
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
HELP EASY MATH PROBLEMS
anastassius [24]

Answer:

21. 98 cents

Gabby placed in 4th.

Step-by-step explanation:

Assuming Steven is paying, 51+45+2=98 cents.

6 0
3 years ago
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What is the surface area of a rectangular prisim with a height of 22 and area of 64
NikAS [45]

Answer:

1408 multiply 64 by 22 for a total of 1,408

6 0
3 years ago
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