find the missing side of each triangle. round your answers to the nearest tenth if necessary.
32)
33)
34) two sides of a triangle have the following measures, 6 and 17. find the range of possible measures for the third side.
a) 6

Question

find the missing side of each triangle. round your answers to the nearest tenth if necessary.
32)
33)
34) two sides of a triangle have the following measures, 6 and 17. find the range of possible measures for the third side.
a) 6 < x < 23  b) -11 < x < 11
c) 11 < x < 23  d) 11 < x < 17
35) which equation represents a line that is parallel to the line y=1/2x-3
a) y=2x-3  b) y=1/2x-2
c) y= -1/2x-3  d) y=2x+4
36) which equation represents a line that is perpendicular to the line y= -1/4x+5.
a) y=4x-2  b) y= -1/4x+4
c) y=1/4x-6  d) y=-4x+5
37) what is the equation for a perpendicular line to the previous problem.
a) y=(2/3)x-1  b) y=(-3/2)x+1
c) y=(3/2)x-1  d) y=(-2/3)x+1
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Accepted Answer

### 32)
# Explanation:
## Step1: Identify triangle type
It's a right triangle, so use Pythagorean theorem: $a^2 + b^2 = c^2$, where $c = 13$, $b = 12$, find $a = x$.
## Step2: Apply Pythagorean theorem
$x^2 + 12^2 = 13^2$
$x^2 + 144 = 169$
$x^2 = 169 - 144 = 25$
$x = \sqrt{25} = 5$
# Answer: $5$ m

### 33)
# Explanation:
## Step1: Identify triangle type
Right triangle, use Pythagorean theorem: $a^2 + b^2 = c^2$, where $a = 12$, $b = 5$, find $c = x$.
## Step2: Apply Pythagorean theorem
$x^2 = 12^2 + 5^2$
$x^2 = 144 + 25 = 169$
$x = \sqrt{169} = 13$
# Answer: $13$ mi

### 34)
# Explanation:
## Step1: Recall triangle inequality
For a triangle with sides $a$, $b$, $c$, $|a - b| < c < a + b$. Here $a = 17$, $b = 6$.
## Step2: Calculate range
$|17 - 6| < x < 17 + 6$
$11 < x < 23$
# Answer: C) $11 < x < 23$

### 35)
# Explanation:
## Step1: Recall parallel lines rule
Parallel lines have equal slopes. The given line $y = \frac{1}{2}x - 3$ has slope $\frac{1}{2}$.
## Step2: Check options
Option B: $y = \frac{1}{2}x - 2$ has slope $\frac{1}{2}$, same as given.
# Answer: B) $y = \frac{1}{2}x - 2$

### 36)
# Explanation:
## Step1: Recall perpendicular lines rule
Perpendicular lines have slopes that are negative reciprocals. Given slope: $-\frac{1}{4}$, negative reciprocal is $4$.
## Step2: Check options
Option A: $y = 4x - 2$ has slope $4$, which is negative reciprocal of $-\frac{1}{4}$.
# Answer: A) $y = 4x - 2$

### 37)
(Assuming "previous problem" is 36, line $y = -\frac{1}{4}x + 5$)
# Explanation:
## Step1: Recall perpendicular slope
Slope of given line: $-\frac{1}{4}$, negative reciprocal is $4$? Wait, no—wait, 36's line is $y = -\frac{1}{4}x + 5$, so perpendicular slope is $4$? Wait, no, maybe "previous" is 35? Wait, 35's line is $y = \frac{1}{2}x - 3$, perpendicular slope would be $-2$? No, the options have $\frac{2}{3}$, $-\frac{3}{2}$, etc. Wait, maybe typo, but assuming 36's line: no, 37's options have slopes $\frac{2}{3}$, $-\frac{3}{2}$, $\frac{3}{2}$, $-\frac{2}{3}$. Wait, maybe "previous" is a line with slope $\frac{2}{3}$? No, re - check. Wait, maybe the "previous problem" is a line with slope $\frac{2}{3}$? No, the options suggest perpendicular to a line with slope $\frac{2}{3}$ (since negative reciprocal of $\frac{2}{3}$ is $-\frac{3}{2}$, but option B is $y = -\frac{3}{2}x + 1$? Wait, no—wait, if the previous line (say, from a prior problem) had slope $\frac{2}{3}$, perpendicular slope is $-\frac{3}{2}$. But the options: B is $y = -\frac{3}{2}x + 1$, D is $y = -\frac{2}{3}x + 1$. Wait, maybe the previous line was $y=\frac{2}{3}x - 1$ (option A of 37's previous? No, 37's options are for its own previous. Wait, maybe the problem has a typo, but following the options: if the line to be perpendicular to has slope $\frac{2}{3}$, then perpendicular slope is $-\frac{3}{2}$ (option B). But assuming the previous problem (say, a line with slope $\frac{2}{3}$), then:
## Step1: Perpendicular slope
If a line has slope $m$, perpendicular slope is $-\frac{1}{m}$. If $m=\frac{2}{3}$, then $-\frac{3}{2}$.
## Step2: Check options
Option B: $y = -\frac{3}{2}x + 1$ has slope $-\frac{3}{2}$, which is perpendicular to slope $\frac{2}{3}$.
# Answer: B) $y = -\frac{3}{2}x + 1$ (assuming previous line had slope $\frac{2}{3}$)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

32)

Step1: Identify triangle type

Step2: Apply Pythagorean theorem

Step1: Identify triangle type

Step2: Apply Pythagorean theorem

Step1: Recall triangle inequality

Step2: Calculate range

Answer:

33)