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Papessa [141]
3 years ago
8

I need help with this problem please!!

Mathematics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

1) 0.1 cents per ounce

2) 0.1 cents per ounce

Step-by-step explanation:

1) 4.64$/74 ounces= 0.0627027027 cents per ounce

2) 2.91$/42 ounces= 0.06928571428 cents per ounce

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HERE ARE SOME LOVELY POINTS !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASE HELP KE
Goshia [24]

Answer:

Aww spanx! :) You are very kind!

Step-by-step explanation:

4 0
3 years ago
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The volume of a right cone is 264 π units. if its circumference measures 12 π units, find its height
abruzzese [7]

Answer:

r = \frac{12\pi }{\pi } / 2 = 6

volume = 1/3 * π * r² * h

h = 3 * V/π * r^{2}

h = \frac{3 * 264\pi }{\pi  * 6^{2} } = 22

8 0
3 years ago
The volume of water and a rectangle swimming pool can be modeled by the function v(x)=x^3+13x-210. If the depth of the pool is g
S_A_V [24]

Answer:

Required,

L=\frac{3+\sqrt{37}}{2}-\frac{144}{x-3}

W=\frac{3-\sqrt{37}}{2}-\frac{144}{x-3}

Step-by-step explanation:

Given volume and depth respectively,

V(x)=x^3+13x-210 and x-3

To find length and width of the rectanglular swiming pool we know,

Volume=length\timesheight\timesdepth.

Let depth=D=x-3, length=L, width=W, then

V=DLW

x^3+13x-210= LW(x-3)

LW=\frac{x^3+13x-210}{x-3}

After divide we will get x^2-3x-22 with remainder -144.

Thus,

x^3+13x-210=(x^2-3x-22)(x-3)-144=(x-3)LW

Now to find root of,

x^2-3x-144=\frac{3\pm\sqrt{9+88}}{2}=\frac{3\pm \sqrt{37}}{2}

Thus,

L=\frac{3+\sqrt{37}}{2}-\frac{144}{x-3}

W=\frac{3-\sqrt{37}}{2}-\frac{144}{x-3}

5 0
3 years ago
Find the surface area of the prism. 4 cm 3 cm 5 cm 9 cm
iren [92.7K]

Answer:

540cm^3

Step-by-step explanation:

4*3*5*9=12*5*9=60*9=540

8 0
4 years ago
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A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distance s​ (in feet) of the ball from th
ra1l [238]
Hello,

s=48t-16t²

a)
s=0==>16t(3-t)=0==>t=0 or t=3

b)
s>32==>48t-16t²>32===>16(t²-3t+2)<0
Δ=9-8=1
==>16(t-1)(t-2)<0

==>1<t<2 (negative between the roots)

5 0
3 years ago
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