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galben [10]
3 years ago
12

Find BCAC = 7BD = 7.5DE = 1.5AE = 11​

Mathematics
1 answer:
tatuchka [14]3 years ago
3 0

5

CE=AE-AC=4

CD=CE-DE=4-1.5=2.5

BC=BD-CD=7.5-2.5=5

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Which are sums of perfect cubes? Check all that apply. 8x6 27 x9 1 81x3 16x6 x6 x3 27x9 x12 9x3 27x9.
Black_prince [1.1K]

The equations which are sums of perfect cubes are as follows;

\rm 8x^6+27x^9+1

\rm x^6+x^3

\rm 27x^9+x^{12}

<h3>Perfect cubes;</h3>

Perfect cubes are the numbers that are the triple product of the same number.

We have to determine

Which are sums of perfect cubes?

1. The given equation is \rm 8x^6+27x^9+1.

The equation can be written as;

\rm 8x^6+27x^9+1\\\\2^3(x^2)^3+3^3(x^3)^3+1^3

All the terms in the expression are can be represented as perfect cubes.

2.  The given equation is \rm 81x^3+16x^6.

The equation can be written as;

\rm 81x^3+16x^6\\\\9^2x^3+4^2(x^2)3

In this, all the terms are not perfect cubes, some of them are squares.

3. The given equation is \rm x^6+x^3.

The equation can be written as;

\rm x^6+x^3\\\\(x^2)^3+x^3

All the terms in the expression are perfect cubes.

4.  The given equation is \rm 27x^9+x^{12}.

The equation can be written as;

\rm 27x^9+x^{12}\\\\3^3(x^3)^3+(x^4)^3

All the terms in the expression are perfect cubes.

5.   The given equation is \rm 9x^3+27x^9.

The equation can be written as;

\rm 9x^3+27x^9\\\\3^2x^3+3^3(x^3)^3

In this, all the terms are not perfect cubes, some of them are squares too.

To know more about perfect cubes click the link given below.

brainly.com/question/4701925

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2 years ago
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Answer:

centre (5, 6 ) , r = \sqrt{37}

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

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x² - 10x + y² - 12y = - 24

using the method of completing the square

add ( half the coefficient of the x / y terms)² to both sides

x² + 2(- 5)x + 25 + y² + 2(- 6)y + 36 = - 24 + 25 + 36

(x - 5)² + (y - 6)² = 37 ← in standard form

with centre (h, k ) = (5, 6 ) and r = \sqrt{37}

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