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vlada-n [284]
3 years ago
9

a recipe calls for 5 tablespoons of orange juice a can of orange juice is 11 fluid ounces how many tablespoons of juice are in a

can​
Mathematics
1 answer:
BartSMP [9]3 years ago
6 0

Answer: 66 teaspoons

Step-by-step explanation: To get the answer all you have to do is multiply it by 6. 11 oz x 6 teaspoons = 66 teaspoons

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Charra [1.4K]
The answer is Y=7/4
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3 years ago
Can somebody please help me​
Westkost [7]
-5 = n + 14
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PLEASE, I NEED HELP! I will not accept nonsense answers, but I will give the person a BRAINLIEST if you get it correct.
alex41 [277]

Answer:  The last choice is correct. Edna can score 5 times as many points in the next level as in the level she has reached

Step-by-step explanation:    chart of values  for y= 5^{x}  

x is the level   y, points possible)

x    y

1     5

2   25

3   125

4    625

5    3125

You can see the exponential pattern

For what it's worth, two views of the graph of the equation are attached

The point values are astronomical!

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3 years ago
Solve the inequalities for x, and match each solution to its number line graph.
rusak2 [61]

Answer:

Do you still need help??

Step-by-step explanation:

3 0
2 years ago
The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

y = 2-x \,\,\,\,\,,  1 \leq x \leq 2.

If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

7 0
3 years ago
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