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VladimirAG [237]
3 years ago
12

If M (4.2) is the midpoint of AB and A (-5, -3), find the coordinates of B.

Mathematics
1 answer:
loris [4]3 years ago
4 0

Given:

M(4,2) is the midpoint of AB and A(-5, -3).

To find:

The coordinates of B.

Solution:

Let the coordinates of point B are (a,b).

Midpoint formula:

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

M(4,2) is the midpoint of AB and A(-5, -3).

M=\left(\dfrac{-5+a}{2},\dfrac{-3+b}{2}\right)

(4,2)=\left(\dfrac{-5+a}{2},\dfrac{-3+b}{2}\right)

On comparing both sides, we get

4=\dfrac{-5+a}{2}

8=-5+a

8+5=a

13=a

And,

2=\dfrac{-3+b}{2}

4=-3+b

4+3=b

7=b

Therefore, the coordinates of point B are (13,7).

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From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
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To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12

Source:
http://www.1728.org/quadr4.htm




5 0
3 years ago
Compare the graph of g(x) = 6x2 with the graph of f(x) = x2.
sertanlavr [38]

Answer:

see explanation

Explanation:

the graph of g(x) is the graph of f(x) shifted vertically by

+ 6 units

or equivalent to a translation  

(

0

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in general  

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=

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0

shift is  

(

0

a

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↑

⏐

⏐

⏐

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for  

a

<

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shift is  

(

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graph{(y-x^2)(y-x^2-6)=0 [-20, 20, -10, 10]}

Step-by-step explanation:

3 0
3 years ago
PLEASE HELP FAST<br><br> answers:<br><br> a.) 4 <br><br> b.) 10 <br><br> c.) 6 <br><br> d.) 1
VLD [36.1K]

Answer:

4

Step-by-step explanation:

There are 5 points. So there are 4 segments. The segments are the line between two points.

6 0
3 years ago
Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0
2 years ago
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