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n200080 [17]
4 years ago
13

Which of these is the area of a sector of a circle with r = 18”, given that its arc length is 6π?

Mathematics
1 answer:
elena-14-01-66 [18.8K]4 years ago
7 0
I hope this helps you

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What does Juliet do after<br> worrying about whether the<br> potion will work?
galina1969 [7]

Answer:

tdyt8ugt77r6r5etstdd56d6e6d5s5e5e5e6tdufycydycycycycycyxtztxyvib>

8 0
3 years ago
The center of the circle whose equation is (x + 2)² + (y - 3)² = 25 is
joja [24]

Answer:

Step-by-step explanation:

center: (-2,3)

Radius: 5

4 0
3 years ago
Please answer this question​
Tems11 [23]

\bold{\huge{\underline{ Solution }}}

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

• \sf{ Polynomial :- ax^{2} + bx + c }

• The zeroes of the given polynomial are α and β .

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

Here, we have polynomial

\sf{ = ax^{2} + bx + c }

<u>We </u><u>know </u><u>that</u><u>, </u>

Sum of the zeroes of the quadratic polynomial

\sf{ {\alpha} + {\beta} = {\dfrac{-b}{a}}}

<u>And </u>

Product of zeroes

\sf{ {\alpha}{\beta} = {\dfrac{c}{a}}}

<u>Now, we have to find the polynomials having zeroes </u><u>:</u><u>-</u>

\sf{ {\dfrac{{\alpha} + 1 }{{\beta}}} ,{\dfrac{{\beta} + 1 }{{\alpha}}}}

<u>T</u><u>h</u><u>erefore </u><u>,</u>

Sum of the zeroes

\sf{ ( {\alpha} + {\dfrac{1 }{{\beta}}} )+( {\beta}+{\dfrac{1 }{{\alpha}}})}

\sf{ ( {\alpha} + {\beta}) + ( {\dfrac{1}{{\beta}}} +{\dfrac{1 }{{\alpha}}})}

\sf{( {\dfrac{ -b}{a}} ) + {\dfrac{{\alpha}+{\beta}}{{\alpha}{\beta}}}}

\sf{( {\dfrac{ -b}{a}} ) + {\dfrac{-b/a}{c/a}}}

\sf{ {\dfrac{ -b}{a}} + {\dfrac{-b}{c}}}

\bold{{\dfrac{ -bc - ab}{ac}}}

Thus, The sum of the zeroes of the quadratic polynomial are -bc - ab/ac

<h3><u>Now</u><u>, </u></h3>

Product of zeroes

\sf{ ( {\alpha} + {\dfrac{1 }{{\beta}}} ){\times}( {\beta}+{\dfrac{1 }{{\alpha}}})}

\sf{ {\alpha}{\beta} + 1 + 1 + {\dfrac{1}{{\alpha}{\beta}}}}

\sf{ {\alpha}{\beta} + 2 + {\dfrac{1}{{\alpha}{\beta}}}}

\bold{ {\dfrac{c}{a}} + 2 + {\dfrac{ a}{c}}}

Hence, The product of the zeroes are c/a + a/c + 2 .

<u>We </u><u>know </u><u>that</u><u>, </u>

<u>For </u><u>any </u><u>quadratic </u><u>equation</u>

\sf{ x^{2} + ( sum\: of \:zeroes )x + product\:of\: zeroes }

\bold{ x^{2} + ( {\dfrac{ -bc - ab}{ac}} )x + {\dfrac{c}{a}} + 2 + {\dfrac{ a}{c}}}

Hence, The polynomial is x² + (-bc-ab/c)x + c/a + a/c + 2 .

<h3><u>Some </u><u>basic </u><u>information </u><u>:</u><u>-</u></h3>

• Polynomial is algebraic expression which contains coffiecients are variables.

• There are different types of polynomial like linear polynomial , quadratic polynomial , cubic polynomial etc.

• Quadratic polynomials are those polynomials which having highest power of degree as 2 .

• The general form of quadratic equation is ax² + bx + c.

• The quadratic equation can be solved by factorization method, quadratic formula or completing square method.

6 0
2 years ago
It’s geometry please find the length of the missing side
Nesterboy [21]

Answer:

x = 18

Step-by-step explanation:

Missing Components of triangle for angles:

A=39

B=51

C=90

Missing Components of triangle for sides:

a=11.3

b=14

x=18

6 0
3 years ago
a rectangular gift box is 10 inches long, 7 inches wide, 1.25 inches high, and had a surface area of 182.5 square inches. gift w
nalin [4]
1) 2 * (10 * 7 + 7 * 1,25 + 10 * 1,25) = 2 * (70 + 8,75 + 12,5) = 2 * 91,25 = 182,5 square inches - a surface area
2) 182,5 * 0,012 = <span>$</span>2,19 - answer.
6 0
3 years ago
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