Answer:
y = 1/2x - 14
Step-by-step explanation:
Parallel lines have the same gradient
So,
gradient for second line = gradient of the first line = 1/2
Also,
the line passes through (-6,-17)
So, we can find the y-intercept by substitution
y = mx + c
-17 = (1/2 * -6) + c
-17 = -3 + c
c = -14
So,
the equation is y = 1/2x - 14
You are a right triangle. You can tell this by both the 3 sides and the 90 degree angle.
The friction force exerted on the 2400kg car is 9408 Newtons
<h3>
What force will be exerted on the 2400kg car?</h3>
First, we know that the friction force and mass are represented by a proportional relation, this means that we can write:
F = k*M
Where F is the force, M is the mass and k is the constant of proportionality.
We know that for a 1600kg car, a force of 6272N is exerted, replacing that we get:
6272N = k*1600kg
Solving for k we get:
k = (6272N)/(1600 kg) = 3.92 N/kg
Then the proportional relationship is:
F = (3.92 N/kg)*M
So if M = 2400kg, we have:
F = (3.92 N/kg)*2400kg = 9408 N
So the friction force exerted on the 2400kg car is 9408 Newtons
If you want to learn more about proportional realtions:
brainly.com/question/12242745
#SPJ1
Answer: A. ![\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D29%2613%5C%5C13%2610%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The question is asking us to find the product of the matrices. The key difference is the second A has a little <em>T</em> in the exponent. This <em>T</em> means transpose. You multiply A by the transpose of A. To find the transpose, you turn the rows into columns.
![A^T=\left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%263%5C%5C2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now that we have our transpose, we can multiply the matrices.
![\left[\begin{array}{ccc}5&2\\3&-1\\\end{array}\right] \left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right] =\left[\begin{array}{ccc}5*5+2*2&5*3+2(-1)\\3*5+2(-1)&3*3+(-1)(-1)\\\end{array}\right] =\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C3%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%263%5C%5C2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2A5%2B2%2A2%265%2A3%2B2%28-1%29%5C%5C3%2A5%2B2%28-1%29%263%2A3%2B%28-1%29%28-1%29%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D29%2613%5C%5C13%2610%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Tan is sin over cos do tan they = (3/5)/(4/5) = 3/4
