Two technicians are discussing hd-obd for trucks technician a says that the simplest most familiar definition is the diagnostic function of electronic control system to identify or self-diagnose system fault and report fault codes technician b says automotive and light duty truck versions of OBD is called OBD2 who is correct

Answer:

While driving in reverse gear it necessary to have reverse audible alarm , a spotter and a rear view mirror.

Explanation:

It is prefer that to not drive a vehicle in reverse. But if we are driving in reverse gear it is necessary to have a reverse audible alarm , a spotter and a rear view mirror.

Answer:

c

Explanation:

I live in britian

Answer:

sorry im answering questions for the points cuz im built dfferent

Explanation:

Answer: That happens to me too sometimes. I hate when people give the wrong answers...like if u don't know the answer, then don't respond

This question is incomplete, its missing an image which will be uploaded along this Answer.

Answer:

the normal component of force F_n is F((√(r²-s²)) / r)

the tangential component of force F_t is F(s/r)

Explanation:

Given the data in the image;

from the free body diagram, we write the expression for ∅

sin∅ = s/r

cos∅ = (√(r²-s²)) / r

now expression for normal component of force is;

F_n = Fcos∅

we substitute

F_n = F((√(r²-s²)) / r)

Therefore, the normal component of force F_n is F((√(r²-s²)) / r)

Also for force F_t

F_t = Fsin∅

we substitute

F_t = F(s/r)

Therefore, the tangential component of force F_t is F(s/r)

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

[tex]\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}[/tex]

[tex]\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}[/tex]

a = 40

b = 29

Answer:

[tex]y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)[/tex]

Explanation:

Let [tex]y = \sin[\ln(5\cdot x^{2}-2\cdot x)][/tex] and we proceed to find the derivative by the following steps:

1) [tex]y = \sin[\ln(5\cdot x^{2}-2\cdot x)][/tex] Given

2) [tex]y = \sin [\ln[x\cdot (5\cdot x - 2)]][/tex] Distributive property

3) [tex]y = \sin[\ln x + \ln (5\cdot x - 2 )][/tex] [tex]\ln (a\cdot b) = \ln a + \ln b[/tex]

4) [tex]y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)[/tex]  [tex]\frac{d}{dx} (\sin x) = \cos x[/tex]/[tex]\frac{d}{dx}(\ln x) = \frac{1}{x}[/tex]/[tex]\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}[/tex]/Rule of chain/Result