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Sedaia [141]
3 years ago
14

Ramona went to the store with $18.55. After purchasing an item, she has $7.24 left. What was the cost of the

Mathematics
2 answers:
SOVA2 [1]3 years ago
7 0

Answer:

11.31

Step-by-step explanation:

if you subtract 7.24 from 18.55 you will have the answer 11.31

xeze [42]3 years ago
7 0

Answer:

The price of the item was $11.31.

Step-by-step explanation:

You simply subtract the two numbers that are given.

$18.55-$7.24 = $11.31.

And to even check your work, you could add the remaining money, to the solution you got from above.

11.31 + 7.24 gets you your starting amount $18.55.

Hope this helped you! :)

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According to the scale on a map, 1 inch on the map = 15 miles. How many miles are represented by 2.5 inches on the map?
Artyom0805 [142]
2.5 times 15 equals 37.5 miles
7 0
3 years ago
6/100 heueheehhehehhehrhrru
Shalnov [3]

Answer:

0.06

Step-by-step explanation:

6/100 is 0.06

4 0
3 years ago
A pile of sand in the shape of a cone has a diameter of 8 in. and a height of 5 in.
max2010maxim [7]

if it has a diameter of 8, that means its radius is half that, or 4.


\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}~~
\begin{cases}
r=radius\\
h=height\\[-0.5em]
\hrulefill\\
r=4\\
h=5
\end{cases}\implies V=\cfrac{\pi (4)^2(5)}{3}\implies V=\cfrac{80\pi }{3}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hfill \stackrel{using~\pi =3.14}{V= 83.7\overline{3}}~\hfill

8 0
3 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
What is (27/8)^-2/3 ?
Paraphin [41]

Answer:

Step-by-step explanation:

7 0
3 years ago
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