Segment [tex]$s_1$[/tex] has endpoints at [tex]$(3+\sqrt{2},5)$[/tex] and[tex]$(4,7)$[/tex]. Segment [tex]$s_2$[/tex] has endpoints at [tex]$(6-\sqrt{2},3)$[/tex] and[tex]$(3,5)$[/tex]. Find the midpoint of the segment with endpoints at the midpoints of [tex]$s_1$[/tex] and [tex]$s_2$[/tex]. Express your answer as [tex]$(a,b)$[/tex].

Answers

Answer 1

Answer:

The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).

Step-by-step explanation:

Midpoint of a segment:

The coordinates of the midpoint of a segment are the mean of the coordinates of the endpoints of the segment.

Midpoint of s1:

Using the endpoints given in the exercise.

[tex]x = \frac{3 + \sqrt{2} + 4}{2} = \frac{7 + \sqrt{2}}{2}[/tex]

[tex]y = \frac{5 + 7}{2} = \frac{12}{2} = 6[/tex]

Thus:

[tex]M_{s1} = (\frac{7 + \sqrt{2}}{2},6)[/tex]

Midpoint of s2:

[tex]x = \frac{6 - \sqrt{2} + 3}{2} = \frac{9 - \sqrt{2}}{2}[/tex]

[tex]y = \frac{3 + 5}{2} = \frac{8}{2} = 4[/tex]

Thus:

[tex]M_{s2} = (\frac{9 - \sqrt{2}}{2}, 4)[/tex]

Find the midpoint of the segment with endpoints at the midpoints of s1 and s2.

Now the midpoint of the segment with endpoints [tex]M_{s1}[/tex] and [tex]M_{s2}[/tex]. So

[tex]x = \frac{\frac{7 + \sqrt{2}}{2} + \frac{9 - \sqrt{2}}{2}}{2} = \frac{16}{4} = 4[/tex]

[tex]y = \frac{6 + 4}{2} = \frac{10}{2} = 5[/tex]

The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).


Related Questions

When we expand (2x + 1/2)^6, what is the coefficient on the x^4 term?

Answers

Answer: The coefficient before x^4 is 60

Step-by-step explanation:

Hey! So I am not an expert at this, but you have to use the binomial theorem

I have attached of the Pascals Triangle (one shows the row numbering as well)

Basically in a pascal triangle, you add the two numbers above it to get the next number below

As you can see, the rows start from 0 instead of 1

The 6th row contains the numbers 1, 6, 15, 20, 15, 6, 1 which would be the coefficient terms

NOTE: the exponents always add to 6, the first term starts at 6 and decrease it's exponent by 1 each time (6, 5, 4, 3, 2, 1, 0) and the second term increases it's exponent by 1 each time (0, 1, 2, 3, 4, 5, 6)

Using this information the third term from the sixth row (15) would be where it is x^4 (I have circled it on the second image)

It would be 15 × 2^4 × (1/2)^2 = 60

The reason why it is 2^4 and (1/2)^2 is because the third term has the exponents 4 and 2 (bolded on the NOTE) which means that the first term must be put to the power of 4 and the second term must be put to the 2nd power

Sorry for the lousy explanation. I really hope this makes sense! Let me know if this helped :)

Simplify the expression

Answers

Answer:

6

Step-by-step explanation:

3 sqrt(20) / sqrt(5)

We know that sqrt(a) /sqrt(b) = sqrt(a/b)

3 sqrt(20/5)

3 sqrt(4)

3 *2

6

what the person above me said

Express the following repeating decimal as a fraction in simplest form.

Answers

Answer:

[tex]0.\overline{369} = \frac{41}{111}[/tex]

Step-by-step explanation:

x = 0.369369369...

10x = 3.69369369...

100x = 36.9369369...

1000x = 369.369369...

1000x - x = 369

999x = 369

[tex]x = \frac{369}{999} \\\\x = \frac{123}{333}\\\\x = \frac{41}{111}[/tex]

21(2-y)+12y=44 find y​

Answers

Answer: y= -2/9
Explanation:
21(2-y)+12y=44
42-21y+12y=44
42-9y=44
-9y=2
y=-2/9

Answer:

[tex]\textbf{HELLO!!}[/tex]

[tex]21\left(2-y\right)+12y=44[/tex]

[tex]42-21y+12y=44[/tex]

[tex]~add ~similar\:elements[/tex]

[tex]42-9y=44[/tex]

[tex]Subtract~42~from~both~sides[/tex]

[tex]42-9y-42=44-42[/tex]

[tex]-9y=2[/tex]

[tex]Divide\:both\:sides\:by\:}-9[/tex]

[tex]\frac{-9y}{-9}=\frac{2}{-9}[/tex]

[tex]y=-\frac{2}{9}[/tex]

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hope it helps...

have a great day!

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