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Reil [10]
3 years ago
13

Which size ice cream dessert gives you the best price per ounce? Small 6-oz for $2.49. Medium 10-oz cup for $3.49. Large 16-oz c

up for $4.99. Super size 24-oz cup for $7.69.
Mathematics
1 answer:
Arlecino [84]3 years ago
8 0

Medium 10 oz Cup For $3.49 Is The Answer

Hope That Helps!

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the table shows the highest of four students arrange the students in order from shortest to tallest Kim 56.03 + x + 56.3 + 56.14
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56.03 56.14 56.3 57.1
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3 years ago
Consider the numbers √27 and √39.
kobusy [5.1K]
√27 = 5.2 (to 1 dp)
√39 = 6.2 (to 1 dp)
A. √39 is greater.
B. 6 is the only whole number between the two.
Hope this helped :)
3 0
3 years ago
Please help I don't know how to explain
spin [16.1K]
Well, 1/4 = 2x.
1/4 / x = 2

cause 2x is multiplication
and the other side shows divison for the right side of the and

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7 0
3 years ago
Find the general solution of the given differential equation. cos^2(x)sin(x)dy/dx+(cos^3(x))y=1 g
eimsori [14]

If the given differential equation is

\cos^2(x) \sin(x) \dfrac{dy}{dx} + \cos^3(x) y = 1

then multiply both sides by \frac1{\cos^2(x)} :

\sin(x) \dfrac{dy}{dx} + \cos(x) y = \sec^2(x)

The left side is the derivative of a product,

\dfrac{d}{dx}\left[\sin(x)y\right] = \sec^2(x)

Integrate both sides with respect to x, recalling that \frac{d}{dx}\tan(x) = \sec^2(x) :

\displaystyle \int \frac{d}{dx}\left[\sin(x)y\right] \, dx = \int \sec^2(x) \, dx

\sin(x) y = \tan(x) + C

Solve for y :

\boxed{y = \sec(x) + C \csc(x)}which follows from [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}.

7 0
2 years ago
On a trip, a motorist drove 150 miles in the morning and 50 miles in the afternoon. His average rate in the morning was twice hi
Lina20 [59]

Answer:

Morning's average rate = 50 mph, and Afternoon's average rate = 25 mph.

Step-by-step explanation:

Suppose he drove 150 miles for X hours, then his average rate in the morning was (150/X) mph.

Given that he spent 5 hours in driving.

And he drove 50 miles for (5-X) hours, then his average rate in the afternoon was 50/(5-X) mph.

Given that his average rate in the morning was twice his average rate in the afternoon.

(150/x) = 2 * 50/(5-x)

150/x = 100/(5-x)

Cross multiplying terms, we get:-

150*(5-x) = 100*x

750 - 150x = 100x

750 = 100x + 150x

750 = 250x

x = 750/250 = 3.

It means he spent 3 hours in the morning and 2 hours in the afternoon.

So morning's average rate = 150/3 = 50 mph.

and afternoon's average rate = 50/(5-3) = 25 mph.

8 0
3 years ago
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