x² - 4x - 12 = 0
Factor the left side: (x + 2) · (x - 6) = 0
The equation is true if either factor is zero.
If (x + 2) = 0 then x = -2 .
If (x - 6) = 0 then x = 6 .
Answer:
a) weight of the car = 2816,1 lbs
b) 2773 lbs
Step-by-step explanation:
The equilibrium force is 490 lbs. That force keep the car at rest, then
∑ Fy = 0 and ∑Fx = 0
Forces acting on the car:
The external force 490 lbs
weight of the car uknown
Normal force
sin∠10° = 0,174
cos∠10° = 0.985
∑Fx = 0 mg*sin10°- 490 = 0 ∑Fy = 0 mg*cos10° - N = 0
mg*0,174= 490
mg = 490 / 0,174
mg = 2816,1 lbs
weight of the car = 2816,1 lbs
The Normal force
mg*cos10° - N = 0 2816,1 * 0,985 = N
N = 2773 lbs
Then equal force in magnitude and in opposite direction will car exets on the driveway
Alternate exterior angles because the are OUTSIDE of the parallel lines
If you're looking for the number of dogs (I'm assuming, you didn't specify), all you'd have to do is divide the number by two, then round down to find the number of dogs, then check your answer by adding that number and that number plus 5.
For example, yours would look like 125/6=62.5, and rounding down to 60.
Then check your answer by adding 60 and 65, the number of cats. 60+65=125.
There are 60 dogs inside the kennel.
Answer:
a. P(male) = 0.4
b. P(no sport and male) = 0.1
c. Unclear question (isn't it the same as b.?)
Step-by-step explanation:
The data below is what I've worked according to, which isn't very clear from the question so the answers are only correct if this is the correct table of data;
![\left[\begin{array}{ccc}&No \ Sports&Sports\\Female&10&32\\Male&7&21\end {array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%26No%20%5C%20Sports%26Sports%5C%5CFemale%2610%2632%5C%5CMale%267%2621%5Cend%20%7Barray%7D%5Cright%5D)
a.
b.
Using the tree diagram in the picture;
