I believe the answer is D because the pathagreom theorum proves it true (excuse my terrible grammar)
It is given in the question that,
Since we have the value of r given, so we have to use the formula to find the nth term of the geometric progression, which is
Substituting the values of a and r, we will get
So the correct option is the third option .
Answer: If we define 2:00pm as our 0 in time; then:
at t= 0. the velocity is 30 mi/h.
then at t = 10m (or 1/6 hours) the velocity is 50mi/h
Then, if we think in the "mean acceleration" as the slope between the two velocities, we can find the slope as:
a= (y2 - y1)/(x2 - x1) = (50 mi/h - 30 mi/h)/(1/6h - 0h) = 20*6mi/(h*h) = 120mi/
Now, this is the slope of the mean acceleration between t= 0h and t = 1/6h, then we can use the mean value theorem; who says that if F is a differentiable function on the interval (a,b), then exist at least one point c between a and b where F'(c) = (F(b) - F(a))/(b - a)
So if v is differentiable, then there is a time T between 0h and 1/6h where v(T) = 120mi/
Answer:
Step-by-step explanation:
Multiply each term of the first polynomial with the second polynomial. Then combine the like terms.
(3a<em>² + 5a - 2)* (5a² -3a + 4)</em>
<em> = 3a² *(5a² -3a + 4) + 5a*(5a² -3a + 4) - 2*(5a² -3a + 4)</em>
<em>=3a²*5a² - 3a*3a² + 4*3a² + 5a*5a² - 3a*5a + 4*5a + 5a²*(-2) - 3a*(-2) + 4*(-2)</em>
<em>=15a⁴ - 9a³ + 12a² + 25a³ - 15a² + 20a - 10a² + 6a - 8</em>
<em>= 15a⁴ </em><u><em>- 9a³ + 25a³</em></u><em> +</em><u><em> 12a² - 15a² - 10a²</em></u><em> +</em><u><em> 20a +6a </em></u><em>- 8</em>
<em>= 15a⁴ + 16a³ - 13a² +26a - 8</em>
Split the pie into 3 sections and shade only one of them