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3 years ago

14 POINTS 14 Points 14 points

Mathematics

2 answers:

viva [34]3 years ago

8 0

Answer:

r = 10 cm

Step-by-step explanation:

r² = 8² + 6²

r² = 64 + 36 = 100

r = 10 cm

Luden [163]3 years ago

7 0

<h2>Answer:</h2>

<u>By pythagorean property</u>,

r² = 6² + 8²

r² = 36 + 64

r² = 100

r = √100

r = 10cm.

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Step-by-step explanation:

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2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

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Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

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$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

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$9x^{2}$ $-3x$ $-210$

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$42x$ $-210$

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5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

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Circumference - Item 2838 Question 4 of 7

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Since we are given the diameter, it’s easier to use the first formula, π*d. Let’s not be too hard on ourselves.

Okay, so let’s state our decided formula:
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