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1 year ago

5(b)

Mathematics

1 answer:

viva [34]1 year ago

7 0

If the side length is greater than 11.11 cm then it will not overflow.

Otherwise, it will overflow.

If Joe tips the bucket of water in a cuboid container and the water is not overflowing then the cuboid container must be of volume greater than 1370 cm³.

We find the cube root of 1370 cm³.

$\sqrt[3]{1370} \approx11.11$

Then the cuboid container should have a side of length greater than 11.11 cm.

Here the statement "If I tip my bucket of water in the cuboid container, it will never overflow" is correct or wrong based on the information that the container has a side length lesser or greater than 11.11 cm.

If the side length is greater than 11.11 cm then it will not overflow.

Otherwise, it will overflow.

Learn more about volume here-

brainly.com/question/1578538

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I will repot any links

asambeis [7]

Answer:

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Step-by-step explanation:

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2 years ago

What is the approximate area of a circle enclosed by a piece of rope 50.24 inches long? (Use the fact that π ≈ 3.14 to make your

Nikitich [7]

Answer:

the approximate area of this circle is 200.96 inches long.

Step-by-step explanation:

To answer this problem we need to remember that the area of a circle is given by the formula:

Area = π $r^2$ where r is the radius.

and the perimeter is:

Perimeter = 2πr

Now, the problem tells us that the circle is enclosed by a piece of rope that's 50.24 inches long. So the perimeter of the circle is 50.24 inches.

Since we have the value of the perimeter and the value of pi, we are going to substitute these values in the perimeter formula to find r.

Perimeter = 2πr

50.24=2(3.14)r

50.24= 6.28r

50.24/6.28= r

8= r

Thus, the radius of the circle is 8 inches long.

Now, we can use this value to find the area of the circle:

Area = π $r^2$

Area = π $8^2$

Area = 3.14 (64)

Area = 200.96

Therefore, the approximate area of this circle is 200.96 inches long.

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2 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.