It is a very simple answer but it works!
Answer:
6.8yards/seconds
Step-by-step explanation:
Speed = Distance/Time
Given the following
Distance covered = 1232 yards
Time taken = 180 seconds
Substitute to get the speed
Speed = 1232/180
Speed = 6.84
Hence whirlwind's speed in yards per seconds to the nearest tenth is 6.8yards/seconds
By using definition of proportionality and Pythagorean theorem, the values of x and y associated with the geometrical system formed by two right triangles are 36 / 5 and 104 / 5, respectively.
<h3>How to analyze a geometrical system formed by two proportional right triangles</h3>
Herein we find a picture of a geometrical system formed by two proportional right triangles with two variables: {x, y} These variables can be found by using the definition of proportionality and the Pythagorean theorem:
5 / 8 = 12 / (12 + x) = 13 / y
Then, we proceed to solve the system formed by two equations:
(12 + x) / 12 = 8 / 5
12 + x = 96 / 5
x = 96 / 5 - 12
x = 36 / 5
y / 13 = 8 / 5
y = 104 / 5
By using definition of proportionality and Pythagorean theorem, the values of x and y associated with the geometrical system formed by two right triangles are 36 / 5 and 104 / 5, respectively.
To learn more on right triangles: brainly.com/question/6322314
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Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
Answer:
Down below
Step-by-step explanation:
a. The range of y = sinθ is [-1,1]
b. The period of y = cosθ is 2π
c. The asymptotes of y = tanθ are -π2, π2, πn
d. The amplitude of y = sinθ is 1
e. The period of y = tanθ is π
f. The max value of y = cosθ is 1
