Answer:
i’m pretty sure it should be 49..?
Answer:
15. x = 10 3/11
16. ∠Q = 75°
Step-by-step explanation:
<h3>15.</h3>
The two base angles of the isosceles triangle are congruent, so both are 180°-8x. The sum of those, together with the apex angle, will be 180°.
... 2(180 -8x) + (5x -67) = 180
... 360 -16x +5x -67 = 180 . . . . . eliminate parentheses
... -11x = -113 . . . . . subtract 293
... x = 113/11 = 10 3/11
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<em>Note on the solution</em>
This value of x makes the interior base angles of the triangle be 97 9/11°. The apex angle is then -15 7/11°. The solution is correct—all the angles add up properly—but the geometry cannot exist.
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<h3>16.</h3>
The approach is the same. The sum of the two congruent base angles and the apex angle is 180°.
... 2(2x+15) + x = 180
... 5x + 30 = 180
... x = (180 -30)/5 = 30
... (2x +15)° = (2·30 +15)° = 75° = angle Q
Answer:
7. 130 + 150 = 280
360 - 280 = 80 degrees
Step-by-step explanation:
Answer:
3,5,9
Step-by-step explanation:
I use the method
if it is even can be divisible by 2
if the first 3 numbers add to something divisible by 3
4 if the first 2 numbers add up to be divisible by 4
5 has to end in 0 or 5
6 has to be divisible by 2 AND 3
9 has to be divisible by 3
<span>Don't forget S is measured in thousands of units so you are solving for :
100 < 74.5 + 43.75Sin(πt/6)
25.5 < 43.75Sin(πt/6)
Sin(πt/6) >25.5/43.75 = 0.582857
ASrcSin(πt/6) > 0.62224 radians
πt/6 > 0.62224
t > 6 x 0.62224/π = 1.1884 (4dp)
This initial value occurs when the sine value is increasing and it will reach its maximum value of 1 when Sin(πt/6) = Sinπ/2, that is when t = 3.
Consequently, monthly sales exceed 100,000 during the period between t = 1.1884 and 4.8116
[3 - 1.1884 = 1.8116 so the other extreme occurs at 3 + 1.8116]
Note : on the basis of these calculations, January is 0 ≤ t < 1 : February is 1 ≤ t < 2 :....May is 4 ≤ t < 5
So the period when sales exceed 100,000 occurs between Feb 6 and May 25 and annually thereafter.</span>
