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77julia77 [94]
2 years ago
12

Q8 The midpoint of GH is M (2,2). One endpoint is H (1,9). Find the coordinates of endpoint G.

Mathematics
1 answer:
Amanda [17]2 years ago
4 0

Answer:

(3,-5)

Step-by-step explanation:

Given

M = (2,2)

H = (1,9)

Required

Determine G

This is calculated using the following midpoint formula;

M(x,y) = (\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})

Where

(x,y) = (2,2) and (x_1,y_1) = (1,9)

Substitute these values in the formula above

(2,2) = (\frac{1 + x_2}{2},\frac{9 + y_2}{2})

Solving for x_2

2 =\frac{1 + x_2}{2}

Multiply both sides by 2

2 * 2 = 1 + x_2

4 = 1 + x_2

x_2 = 4 - 1

x_2 = 3

Solving for y_2

2 = \frac{9 + y_2}{2}

Multiply both sides by 2

2 * 2 = 9 + y_2

4 = 9 + y_2

y_2 = 4 - 9

y_2= -5

Hence, the coordinates of G is (3,-5)

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Answer:

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Step-by-step explanation:

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7x + 2 + 10x - 9 = 180

17x - 7 = 180 ( add 7 to both sides )

17x = 187 ( divide both sides by 17 )

x = 11 → d

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Answer:

Part 1) The length of two sides and the measure of the included angle (Side-Angle-Side)

Part 2) b^2=a^2+c^2-2(a)(c)cos(B)  

Part 3) b=11.6\ in

Step-by-step explanation:

we have

In the triangle ABC

a=11\ in\\c=9\ in\\B=70^o

Part 1) Which information about the triangle is given?

In this problem we have the length of two sides and the measure of the included angle (Side-Angle-Side)

see the attached figure to better understand the problem

Part 2) Which formula can you use ti find b?

I can use the law of cosines

b^2=a^2+c^2-2(a)(c)cos(B)  

we have

a=11\ in\\c=9\ in\\B=70^o

substitute the given values

b^2=11^2+9^2-2(11)(9)cos(70^o)

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b=11.59\ in

Part 3) What is b, rounded to the nearest tenth?

Remember that

To Round a number

a) Decide which is the last digit to keep  

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c) But increase it by  1 if the next digit is  5 or more (this is called rounding up)

In this problem we have

11.59\ in

We want to keep the digit 5

The next digit is 9 which is 5 or more, so increase the "5" by 1 to "6"

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Step-by-step explanation:

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igomit [66]

Answer:

Vertical A @ x=3 and x=1

Horizontal A nowhere since degree on top is higher than degree on bottom

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Step-by-step explanation:

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I'm going to factor the bottom first:  (x-3)(x-1)

So we have possible vertical asymptotes at x=3 and at x=1

To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)

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

x^2-4x+3 |      x^3-5x^2+4x-25

                  - ( x^3-4x^2+3x)

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

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                       -   (-x^2+4x-3)

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

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So the slant asymptote is to x-1

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