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Yuri [45]
3 years ago
5

Help me with this please!!

Mathematics
1 answer:
valina [46]3 years ago
3 0
<h2><u>Given:</u><u>-</u></h2>

  • Points C = (-7,2) → \sf{(X_1,Y_1)}
  • D = (3,12) → \sf{(X_2,Y_2)}

<h2><u>To </u><u>Find</u><u>:</u><u>-</u></h2>

  • The Midpoint of CD.

<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

Let,

Midpoint of CD be (x,y).

WKT,

\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}

→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}

→\;{\sf{\frac{-4}{2},\frac{14}{2}}}

→\;{\sf{-2,7}}

The Midpoint of CD ◕➜ \Large{\red{\mathfrak{(-2,7)}}}

Let,

The centre be O

Radius = CO & OD

Here, C = (-7,2) → \sf{(X_1,Y_1)}

O = (-2,7) → \sf{(X_2,Y_2)}

\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}

→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}

→\;{\sf{\sqrt{5²+5²}}}

→\;{\sf{\sqrt{25+25}}}

→\;{\sf{\sqrt{50}}}

→\;{\sf{5√2 (or) 7.07}}

Radius of Circle ◕➜ \Large{\red{\mathfrak{7.07}}}

<h2>Option D.</h2>

Hope It Helps You ✌️

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5 0
3 years ago
Help, please and thanks!
azamat
B.
X-17=-5
Add 17 to both sides

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7 0
3 years ago
Solve the following by using mathematical Induction. For &gt;/ 1
cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)

1. When n=1, we have

  • in left part 1^2=1;
  • in right part \dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.

2. Assume that for all k following equality is true

1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)

3. Prove that for k+1 the following equality is true too.

1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)

Consider left part:

1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}

Consider right part:

\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)

We get the same left and right parts, so the equality is true for k+1.

By mathematical induction, this equality is true for all n.

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