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patriot [66]
3 years ago
9

The biggest bowl of mashed potatoes ever made weighed 4.56 x 10^4 pounds and

Mathematics
1 answer:
Bess [88]3 years ago
6 0

We have been given that the biggest bowl of mashed potatoes ever made weighed 4.5\times 10^4 pounds and  was created at the Iowo State Fair in 2005. The biggest deep dish pizza ever  made was in Chicago in 1998. It weighed a whopping 1.23\times 10^5 pounds.

To find how much pizza weighs more than bowl of mashed potatoes, we will subtract weight of mashed potatoes from weight of pizza.

1.23\times 10^5-4.56\times 10^4

1.23\times 100000-4.56\times 10000

123000-45600=77400

Now we will convert 77400 into scientific notation as:

77400=7.74\times 10^4

Therefore, the pizza weighs 77400=7.74\times 10^4 pounds more than the bowl of mashed potatoes.

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16/21 divided by 4/9
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16/21 divided by 4/9 is equivalent to 16/21 x 9/4 = 144/84 which reduces to 12/7. So 12/7 is your answer.
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What is the equivalent expression for 2/3÷4/5
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2/3÷4/5
2/3*5/4
10/12
5/6

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Deb has a board that measures 5 feet in length how many 1/4 Foot long pieces can deb cut from the board.
uranmaximum [27]
Deb can cut 20 pieces from the board
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your cell phone company started a rewards club . for every three texts sent, you get 15 points. you need 1800 points for a price
yanalaym [24]

Answer:

360 texts

Step-by-step explanation:

You get 15 points for each 3 texts.

That means you get 5 points per text.

1800/5 = 360

Answer: You need to send 360 texts.

8 0
4 years ago
4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

5 0
2 years ago
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