Answer:
B
Step-by-step explanation:
Remark
If you started at W and went all the way around the circle until you wound up back at W, you will have gone 360 degrees. Since the angle measurement that prevents that is 85 degrees, the answer must be 360 - 85 = 275.
Answer:
x=-119
Step-by-step explanation:
23-x=142
-23 -23
-x=119
switch the signs
x=-119
Answer: The mean temperature for the first eight days is 6.5 degrees
Step-by-step explanation: The most important piece of clue has been given which is the mean (average) for the observed data set, which is 7 days.
Note that the formula for the mean of a data set is derived as;
Mean = ∑x / f
Where ∑x is the summation of all observed data set and f is the number of data observed, that is 7. The formula now becomes;
6 = ∑x / 7
By cross multiplication, we now have,
6 * 7 = ∑x
42 = ∑x
This means the addition of all temperature observed on the first 7 days is 42. The temperature on the eighth day is now given as 10 degrees, this means the summation of all observed data for the first eight days would become 42 + 10 which equals 52. Therefore when calculating the mean for the first eight days, ∑x is now 52. The formula for the first eight days therefore is derived as follows;
Mean = ∑x / 8
Mean = 52 / 8
Mean = 6.5
The calculations therefore show that the mean temperature for the first eight days in January is 6.5 degrees
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°.
