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2 years ago

Need answers ASAP

1 answer:

6 0

We know, Volume of a Sphere = 4/3 πr³
v = 4/3 * 3.14 * 18³
v = 4.186 * 5832
v = 24,416.64 cm³

Volume of a Sphere = 4/3 πr³
v = 4/3 * 3.14 * 12³
v = 4.186 * 1728
v = 7234.56 cm³

Difference = 24,416.64 - 7234.56 = 17182.08

In short, Your Answer would be: 17182.08 cm³

Hope this helps!

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Isabelle is using a square piece of cardboard to make a box without a lid. She cuts 2-inch squares from each corner and folds up

katrin2010 [14]

<em>The dimensions of the cardboard to make a box without a lid is 11 inches by 11 inches.</em>

<h2>Explanation:</h2>

In this problem, we have that Isabelle is using a square piece of cardboard to make a box without a lid. She cuts 2-inch squares from each corner and folds up the edges. The figure below shows this statement. The finished box has a volume (V) of 98 cubic inches, in other words:

$V=98in^3$

When folding up, we get the box without lip shown in the second figure. So the volume would be:

$V=2xx \\ \\ V=2x^2 \\ \\ But: \\ \\ V=98 \\ \\ Then: \\ \\ 2x^2=98 \\ \\ x^2=49 \\ \\ x=\sqrt{49} \\ \\ x=7in$

Since te cardboard is square, then the side (s) is given by:

$s=2+x+2 \\ \\ s=2+7+2 \\ \\ s=11in$

Therefore:

<em>The dimensions of the cardboard to make a box without a lid is 11 inches by 11 inches.</em>

<h2>Learn more:</h2>

Volume of a box: brainly.com/question/10501080

#LearnWithBrainly

8 0

2 years ago

Read 2 more answers

Can someone check my answer please HELP!

Bezzdna [24]

Let price of one scarf be $x
Price of the other is 3 more than $x
Therefore price of the other scarf =$x +3
The total price she paid = $25.00
That is, x+x+3=25
2x+3=25
2x=25-3
2x=22
x=22/2
x=11
The price of one scarf is $11 and the price of the other is 11+3=$14
The price of the expensive scarf =$14.

6 0

2 years ago

Circle has rad of 1.5m what angle subtended at center of circle by arc of 1m

Mumz [18]

120 degrees or 0.67 radians

5 0

2 years ago

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Natalija [7]

Answer:

The correct answer is A

Step-by-step explanation:

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7 0

1 year ago

Can someone help me with this? PLs i'm so confused!

barxatty [35]

1. E. $sine\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}$

2. L. $cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}$

3. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{5}{12}$

4. Y. $sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}$

5. W. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}$

6. $tan\ B = \frac{b}{a} = \frac{adjacent}{opposite} = \frac{AC}{BC} = \frac{12}{5} = 2\frac{2}{5}$

7. $sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}$

8. W. $cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{2}$

9. I. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

10. $sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}$

11. E. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{1} = \sqrt{3}$

12. I. $tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

13. U. $sin\ A = \frac{a}{c} = \frac{hypotenuse}{opposite} = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}$

14. I. $cos\A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}$

15. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3}$

16. R. $sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{4}{\sqrt{65}} = \frac{4}{\sqrt{65}} * \frac{\sqrt{65}}{\sqrt{65}} = \frac{4\sqrt{65}}{65}$

17. M. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{7}{4} = 1\frac{3}{4}$

18. N. $tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{4}{7}$

19. L. $sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{16}{34} = \frac{8}{17}$

20. H. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \fac{AC}{AB} = \frac{30}{34} = \frac{15}{17}$

21. $tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{16}{30} = \frac{8}{15}$

22. O. $sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

23. O. $cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

24. N. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1$

7 0

2 years ago