Answer:
∠A = 30°
∠B = 60°
∠C= 90°
Step-by-step explanation:
This is a right triangle, you can see it mainly by the red square in C, and it is always used to mark 90 degrees.
Knowing that, you now know <em>∠C is 90°</em>
Now, to find ∠B, you should use the following equation:
This means that the sum of the three angles of a triangle gives 180. ALWAYS. So to find the missing angle, ∠B, do the following:
Fill the values of the equation with the angles you now know:
Solve the equation, passing the 30° and 90° to the other side of the equal sing with Inverse Operation:
<em>B = 60</em>
<em>Hope it helps!!</em>
Answer:
AB = DF (Proved below)
Step-by-step explanation:
Here,
CE = CD + DE [since D is a point over CE]
= DE + EF [ Since, CD = EF (given)]
= DF [ Since, E is a point over DF] ---------------(1)
Again,
AB = CE (given) ------------------------(2)
So from (1) and (2),
AB = CE = DF
⇒ AB = DF (Proved)
You have to estimate the slope of the tangent line to the graph at <em>t</em> = 10 s. To do that, you can use points on the graph very close to <em>t</em> = 10 s, essentially applying the mean value theorem.
The MVT says that for some time <em>t</em> between two fixed instances <em>a</em> and <em>b</em>, one can guarantee that the slope of the secant line through (<em>a</em>, <em>v(a)</em> ) and (<em>b</em>, <em>v(b)</em> ) is equal to the slope of the tangent line through <em>t</em>. In this case, this would be saying that the <em>instantaneous</em> acceleration at <em>t</em> = 10 s is approximately equal to the <em>average</em> acceleration over some interval surrounding <em>t</em> = 10 s. The smaller the interval, the better the approximation.
For instance, the plot suggests that the velocity at <em>t</em> = 9 s is nearly 45 m/s, while the velocity at <em>t</em> = 11 s is nearly 47 m/s. Then the average acceleration over this interval is
(47 m/s - 45 m/s) / (11 s - 9 s) = (2 m/s) / (2 s) = 1 m/s²
Answer:
11.84% approx.
Step-by-step explanation:
Expected return = Respective return × Respective probabilities
= (18.5 × 0.71) + (-7.6 × 0.29)
= 10.931%
Probability Return Probability ×(Return - expected return)²
0.71 18.5 0.71×(18.5 - 10.931)² = 40.67573
0.29 -7.6 0.29×(-7.6 - 10.931)² = 99.585409
Total = 140.261139%
![SD=[\frac{\text{total probability(return-expected return)}^2}{\text{total probability}}]^{(\frac{1}{2})}](https://tex.z-dn.net/?f=SD%3D%5B%5Cfrac%7B%5Ctext%7Btotal%20probability%28return-expected%20return%29%7D%5E2%7D%7B%5Ctext%7Btotal%20probability%7D%7D%5D%5E%7B%28%5Cfrac%7B1%7D%7B2%7D%29%7D)
= ![[\frac{140.261139^2}{140.261139}]^{\frac{1}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7B140.261139%5E2%7D%7B140.261139%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)
= 11.84319 ≈ 11.84% approx.
