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Sati [7]
2 years ago
8

Ivan wants to fill with foam the empty space left in a cube-shaped box after he places a basketball in the box. How many cubic i

nches of foam is needed if the radius of the basketball is 4.75 in. and the sides of the box are each 9 in.? Use 3.14 for π, and round your answer to the nearest whole number. in.3
Mathematics
1 answer:
Ann [662]2 years ago
6 0

Answer:

280 cubic inches

Step-by-step explanation:

Cubic inches of foam required = volume of the box - volume of the basketball

The box has a shape of a cube, so that;

volume of the box = l^{3}

where l is the length of the sides of the box.

volume of the box = 9^{3}

                               = 729 cubic inches

volume of the box = 729 cubic inches

The basketball has the shape of a sphere, so that;

volume of the basketball = \frac{4}{3} \pir^{3}

                                          = \frac{4}{3} x 3.14 x (4.75)^{3}

                                           = 448.693

volume of the basketball = 448.693 cubic inches

Thus,

cubic inches of foam required = 729 - 448.693

                                                   = 280.307

The cubic inches of foam to be used is 280 cubic inches.

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−2x=x^2−6
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Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

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Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

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Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

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The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

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f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

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x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

6 0
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