If you need the answer so go to the google and search for problem
Answer:
Both angles have a measure of 134degrees, y = 27degrees.
Step-by-step explanation:
As per what is given in the problem:
There are 2 parallel lines, both are intersected by a transversal.
Remember the theorem, when two parallel lines are intersected by a transversal, then the alternate exterior angles are congruent.
The is meanse that:
3y + 53 = 7y - 55
Solve using inverse operations:
3y + 53 = 7y - 55
+55 +55
3y + 108 = 7y
-3y -3y
108 = 4y
/4 /4
27 = y
Now, substitute back in to find the value of the angle:
3y + 53
y = 27
3 ( 27 ) + 53
81 + 53
= 134
Since the angles are alternate exterior, they are congruent, hence both angles have a measure of 134degrees.
Answer:
f(5) = 26.672 which is option D
Step-by-step explanation:
From question, f(1) = 2 and f'(x)=√(x^3 + 6)
f(5) = f(1) + (5,1)∫ f'(x) dx
Integrating using the boundary 5 and 1;
f(5) = 2 + (5,1)∫√(x^3 + 6) dx
f(5) = 2 + 24.672
So f(5) = 26.672
Answer:
12.53
Step-by-step explanation:
∠BCA = (180°-40°)/2 = 70°
Laws of Sines: CA/sin 40° = AB/sin 70° = 5/sin 70°
CA = 5/sin 70° * sin 40° = 3.40
area of semicircle CA = (3.1416 * 1.70²) / 2 = 4.54
By Heron's Formula, area of triangle ABC:
1/4 √(a+b+c) (a+b-c) (b+c-a) (c+a-b) = 1/4 * √13.4*3.4*3.4*6.6
= 7.99
Entire region area = 4.54 + 7.99 = 12.53
Answer:
D-the slope of the line will be zero
Step-by-step explanation:
