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Answer:
x² - 3x - 10 = (x - 5) (x + 2)
x² - 3x - 18 = (x + 3) (x - 6)
Step-by-step explanation:
<u>x² - 3x - 10</u>
x² + 2x - 5x - 10
x(x + 2) - 5(x + 2)
(x - 5) (x + 2)
<u>x² - 3x - 18</u>
x² - 6x + 3x - 18
x(x - 6) + 3(x - 6)
(x + 3) (x - 6)
<u>-TheUnknown</u><u>S</u><u>cientist</u>
Answer:

Step-by-step explanation:
Acceleration is second derivative of distance and are related as:

Integrating both sides w.r.to t

Using initial value

We have to calculate the distance covered in time interval [0,5], so:
![\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft](https://tex.z-dn.net/?f=%5Cint%5Climits%5E5_0%20%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cint%5Climits%5E5_0%20%7B%5Cfrac%7B1%7D%7Bt%2B2%7D%7D%20%5C%2C%20dt%5C%5C%5C%5Cs%28t%29%3D%5Bln%7Ct%2B2%7C%5D%5E5_0%5C%5C%5C%5Cs%28t%29%3Dln%7C5%2B2%7C%2Bln%7C0%2B2%7C%5C%5C%5C%5Cs%28t%29%3D%28ln%7C7%7C%2Bln%7C2%7C%29%5C%2Cft)
Answer:
V(X)= 39.10
V(Y)= 40
Step-by-step explanation:
Given that
Total number of student = 140
Bus A - 31
Bus B- 43
Bus C- 27
Bus D- 39
The probability that a student was on the bus is proportional to the number of student. Eg 31/140 in bus A, 43/140 on bus B, ...
E(X) = (31*31/140) + (43*43/140) + (27*27/140) + (39*39/140)
= 35.5

V(X)= 39.10
The bus driver have 1/4 probability on being on any of the buses.
E(Y) = 140/4 = 35

V(Y)= 40
That's a monomial; it has one variable, one coefficient, and one degree.