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lesson 4 - 1 1. if q is between a and m and mq = 7.3 and am = 8.5, then…

Question

lesson 4 - 1

  1. if q is between a and m and mq = 7.3 and am = 8.5, then qa =?

a. 5.8
b. 1.2
c. 7.3
d. 14.6

  1. given: k is between h and j, hk = 2x - 5, kj = 3x + 4, and hj = 24. what is the value of x?

a. 9
b. 5
c. 19
d. 3

  1. if k is the midpoint of \\( \overline { hj } \\), hk = x + 6, and hj = 4x - 6, then kj =?

a. 15
b. 9
c. 4
d. 10

  1. state the segment addition postulate in your own words.
  2. explain what distance along a line means as an undefined term in geometry.

use the number line for items 6 - 9.

  1. what is ab?
  2. what is the coordinate of the midpoint of \\( \overline { ab } \\)? explain how you found your answer.
  3. point m is the midpoint of \\( \overline { ab } \\). what is the coordinate of the midpoint of \\( \overline { am } \\)?
  4. point c is between points a and b. the distance between points b and c is \\( \frac { 1 } { 4 } \\) of ab. what is the coordinate of point c?
  5. \\( \overline { fg } \\) lies on a number line. the coordinate of point f is 8. given that fg = 16, what are the two possible coordinates for point g?

Explanation:

Step1: Find the length of \(QA\)

Since \(Q\) is between \(A\) and \(M\), by the segment - addition postulate \(AM = AQ+MQ\). We know that \(AM = 8.5\) and \(MQ = 7.3\). Then \(AQ=AM - MQ\).
Substitute the values: \(AQ=8.5 - 7.3\)

$$AQ = 1.2$$

Step2: Solve for \(x\) in the second problem

Since \(K\) is between \(H\) and \(J\), by the segment - addition postulate \(HJ=HK + KJ\). We know that \(HK = 2x-5\), \(KJ = 3x + 4\), and \(HJ = 24\).
So, \((2x-5)+(3x + 4)=24\).
First, simplify the left - hand side: \(2x-5+3x + 4=24\), which gives \(5x-1 = 24\).
Add \(1\) to both sides: \(5x=24 + 1=25\).
Divide both sides by \(5\): \(x=\frac{25}{5}=5\)

Step3: Solve for \(x\) and then find \(KJ\) in the third problem

Since \(K\) is the midpoint of \(\overline{HJ}\), \(HK=\frac{1}{2}HJ\). We know that \(HK=x + 6\) and \(HJ = 4x-6\).
So, \(x + 6=\frac{1}{2}(4x-6)\).
Multiply both sides by \(2\): \(2(x + 6)=4x-6\).
Expand: \(2x+12 = 4x-6\).
Subtract \(2x\) from both sides: \(12=4x-6-2x\), which gives \(12 = 2x-6\).
Add \(6\) to both sides: \(2x=12 + 6=18\).
Divide by \(2\): \(x = 9\).
Since \(K\) is the midpoint, \(KJ=HK\). Substitute \(x = 9\) into \(HK=x + 6\), then \(HK=9+6 = 15\), so \(KJ = 15\)

Step4: State the segment - addition postulate

If \(B\) is between \(A\) and \(C\), then \(AC=AB + BC\)

Step5: Explain distance along a line

In geometry, distance along a line (for two points \(A\) and \(B\)) is a non - negative real number that measures the "separation" between the two points. It is calculated as \(d(A,B)=\vert a - b\vert\) where \(a\) and \(b\) are the coordinates of points \(A\) and \(B\) on the number line (or in a coordinate system)

Step6: Find the length of \(AB\)

The coordinate of \(A=-12\) and the coordinate of \(B = 8\). Using the distance formula \(d(A,B)=\vert x_{B}-x_{A}\vert\), \(AB=\vert8-(-12)\vert=\vert8 + 12\vert=20\)

Step7: Find the mid - point of \(\overline{AB}\)

The formula for the mid - point \(M\) of a segment with endpoints \(x_1\) and \(x_2\) is \(M=\frac{x_1 + x_2}{2}\). Here \(x_1=-12\) and \(x_2 = 8\).
\(M=\frac{-12 + 8}{2}=\frac{-4}{2}=-2\)

Step8: Find the mid - point of \(\overline{AM}\)

The coordinate of \(A=-12\) and the coordinate of \(M=-2\) (from step 7). Using the mid - point formula \(M'=\frac{-12+(-2)}{2}=\frac{-14}{2}=-7\)

Step9: Find the coordinate of \(C\)

We know \(AB = 20\) (from step 6). The distance \(BC=\frac{1}{4}AB\), so \(BC=\frac{1}{4}\times20 = 5\).
Since \(B\) has a coordinate of \(8\), and \(C\) is between \(A\) and \(B\), \(C\)’s coordinate is \(8-5 = 3\)

Step10: Find the coordinates of \(G\)

If \(F\) has a coordinate of \(8\) and \(FG = 16\). Using the distance formula \(d(F,G)=\vert x_G-x_F\vert\). Let \(x_F = 8\) and \(d(F,G)=16\). Then \(\vert x_G - 8\vert=16\)
Case 1: \(x_G-8=16\), then \(x_G=16 + 8=24\)
Case 2: \(x_G-8=-16\), then \(x_G=-16 + 8=-8\)

Answer:

  1. B. \(1.2\)
  2. B. \(5\)
  3. A. \(15\)
  4. If \(B\) is between \(A\) and \(C\), then \(AC = AB+BC\)
  5. Distance along a line (for two points) is a non - negative real number calculated as \(d(A,B)=\vert a - b\vert\) where \(a\) and \(b\) are the coordinates of the two points.
  6. \(20\)
  7. \(-2\). Using the mid - point formula \(M=\frac{x_1 + x_2}{2}\) with \(x_1=-12\) and \(x_2 = 8\)
  8. \(-7\)
  9. \(3\)
  10. \(24\) and \(-8\)