1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
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Answer:
x = 7
Step-by-step explanation:
10.5 maps to x with a scale factor of 2/3:
x = 10.5 × 2/3
x = 7
The distributive property:
Answer:
The place where the x and y axis meet, is called the point of origin
Step-by-step explanation:
he data shows the distances, in miles, run by runners in a track club. 6, 3, 8, 8, 6, 6, 8, 2, 5, 2, 5, 10, 3, 5 Create a histog
kakasveta [241]
Answer:
To create a histogram of the given data set, we first must check for the frequency of each value.
6, 3, 8, 8, 6, 6, 8, 2, 5, 2, 5, 10, 3, 5
2 - 2
3 - 2
5 - 3
6 - 3
8 - 3
10 - 1
Now that we have the frequencies of each, we need to add them to each range.
0 - 2 : 2
3 - 5 : 5
6 - 8 : 6
9 - 11 : 1
Refer to the image for the histogram.
