Using subtraction, you would take 3,000 and subtract 1968 from it to leave you 1,032. With that number, the answer to your question is "C"
Ken's distance is: d(t)=1.7t
Ken's shadow is:
(14-6)/d(t)=14/s
8/d(t)=14/s
s=14d(t)/8
s=1.75d(t) and using the value for d(t) we have:
s=1.75(1.7t)
s=2.975t and if you want it in terms of d, d=1.7t, t=d/1.7 so
s=2.975(d/1.7)
s=1.75d
A. x-2y = 4
Is the right answer...
Please mark as the Brainliest...!!!
The trapezoid has two right triangles each with one side of length (13-9)2 = 4/2 = 2.
The other side of each right triangle is such that tan 60° = Heigth / 2
Then height = 2*tan60°.
The area of the trapezoid is height * (base 1 + base 2)/2
Then area = 2*tan (60°) * (13+9)/2 = 38.11
Answer:
∠A=30°
Step-by-step explanation:
let's call the angle at the bottom B and the one next to the 130 C
∠A=3x-6
∠C=180-130= 50°
vertical angles are equal∠B =8x+4
∠A+∠C+∠B=180°
find x:
3x-6+50+8x+4= 180°
11x+48=180°
11x= 132
x=12
find ∠A
∠A=3x-6= 3(12)-6 =30°
