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rodikova [14]
3 years ago
8

The equation x^3 - 6x = 72 has a solution between 4 and 5.

Mathematics
1 answer:
DENIUS [597]3 years ago
5 0

Answer:

4.6

Step-by-step explanation:

X^3-6x=72

Solution between 4 and 5 correct to one decimal place

This means There are 10 Possible solution (4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5.0)

Let x=4.1

X^3-6x=72

=4.1^3-6(4.1)

=68.921-24.6

=44.321

This means 4.1 is far from the answer

Let's try a bigger number

Let x=4.8

X^3-6x=72

4.8^3-6(4.8)

=110.592-28.8

=81.792

Let x=4.7

X^3-6x=72

4.7^3-6(4.7)

=103.823-28.2

=75.623

Let x=4.6

X^3-6x=72

4.6^3-6(4.6)

=97.336-27.6

=69.736

It can be said that the solution is between 4.6 and 4.7

Let x=4.62

X^3-6x=72

4.62^3-6(4.62)

=98.611128-27.72

=70.891128

Let x=4.64

X^3-6x=72

4.64^3-6(4.64)

=99.897344-27.84

=72.057344

Let's=4.65

X^3-6x=72

=4.65^3-6(4.65)

=100.544625-27.9

=72.644625

This is greater than 72

We can conclude that the closest answer is 4.64

Let x=4.64

X^3-6x=72

4.64^3-6(4.64)

=99.897344-27.84

=72.057344

Approximately 72

4.64 to one decimal place=4.6

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Answer:

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Step-by-step explanation:

4 x 6 = 24

Now we have to double those lengths.

We know that 4 x 2 is 8 and 6 x 2 is 12.

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Quadrilateral KMPT is dilated by a scale factor of 3/4 to create K'M'P'T'. The perimeter of quadrilateral KMPT is x units. What
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Answer:

B) \frac{3}{4} x units

Step-by-step explanation:

Let quadrilateral KMPT be a rectangle with dimensions 12 units by 8 units. Then its perimeter would be equal to:

Perimeter of a rectangle = 2 (l + b)

where: l is the length = 12 units and b is the breadth = 8 units. So that:

Perimeter of KMPT = 2 (12 + 8)

                               = 40 units

Dilating KMPT by a scale factor of \frac{3}{4} would create K'M'P'T' of dimensions; \frac{3}{4} × 12 units by \frac{3}{4} × 8 units. Thus, the dimensions of K'M'P'T' would be 9 units by 6 units.

Perimeter of K'M'P'T' = 2 (l + b)

                                  = 2(9 + 6)

                                  = 30 units

Comparing the perimeters of KMPT and K'M'P'T', the perimeter of K'M'P'T' would be  \frac{3}{4} × perimeter of KMPT.

Therefore, if the perimeter of KMPT is x units, then;

         perimeter of K'M'P'T' = \frac{3}{4} * x units

                                            = \frac{3}{4}x units

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Answer:

435.26

Step-by-step explanation:

I hope this is right I looked it up on Google. You search up area of a square pyramid and put your info in.

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