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raketka [301]
4 years ago
7

Target sells 24 bottles of water for $3, and 36 bottles of water for $4. Which is the better buy and by how much?

Mathematics
1 answer:
pishuonlain [190]4 years ago
6 0

Answer:

D) 36 bottles for $4 by 1.4¢ per bottle

Step-by-step explanation:

We calculate bottle per dollar for each option.

#A

Bottles/Dollar=\frac{24\ bottles}{\$3}=8 bottles/dollar

#Both B and C charges the same as A and will have bottle/dollar ration of 8:1

#D

=\frac{36}{4}\\\\=9

Hence, D is the better buy  as it gives you 9bottles per $1 spent as compared to the 8 bottles per $1 spent as in the other 3 options.

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ValentinkaMS [17]

:  \implies \sf 9x + 3 =  - 33

: \implies \sf 9x =  - 33 - 3

:  \implies \sf 9x =  - 36

:  \implies \sf x =  \dfrac{ - 36}{9}  =  - 4

\therefore  \sf x =  -4

8 0
3 years ago
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Is 9/10 larger than 13/14
Reptile [31]

Answer:

13/14 is greater

Step-by-step explanation:

Find the lcm for both:

<u>9 x 7 </u>     = <u>63</u>

10 x 7        70

<u>13 x 5 </u>    = <u>65</u>

14 x 5        70

so therefore 13/14 is greater since it has a greater value when finding the lcm

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3 years ago
Can someone help me, please!!!
kolezko [41]

Answer:

D

The line shows an intercept of -3 and a slope of 5/7. Equation D displays this

Step-by-step explanation:

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B32%7D" id="TexFormula1" title="\sqrt[3]{32}" alt="\sqrt[3]{32}" align="absmid
Elza [17]
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Uranus moves in an elliptical orbit with the sun at one of the foci. The length of the half of the major axis is 2,876,769,540 k
Alina [70]

Answer:

The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

Step-by-step explanation:

Consider the provided information.

The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.

The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).

e=\frac{c}{a}

Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.

0.0444=\frac{c}{2,876,769,540 }

c=127728567.576

Minimum distance of Uranus from the sun is:

a-c=2,876,769,540-127728567.576\\a-c=2749040972.424\approx2749040972

Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

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