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adelina 88 [10]
3 years ago
14

In what ways are driverless cars safer than human drivers?

Mathematics
1 answer:
Kamila [148]3 years ago
4 0

Answer:

Driverless cars may be safer than cars with drivers because they do not get tired or distracted, and they do not engage in risky driving practices such as speeding and/or texting while driving, drinking and driving, etc..

Step-by-step explanation:

Edg. 2020

You might be interested in
8x + 20y = 20<br> 14x + 35y = k<br> For the above system of equations to be consistent, k must equal
Goshia [24]

Answer:

  35

Step-by-step explanation:

The equations will be "consistent" when they describe lines that are not parallel.

<h3>Slopes</h3>

The slope of a line whose equation is written in this form is the opposite of the ratio of the y-coefficient to the x-coefficient:

  line 1 slope = -8/20 = -2/5

  line 2 slope = -14/35 = -2/5

These lines have the same slope, so will be parallel unless they have the same y-intercept.

<h3>Y-intercept</h3>

The y-intercept of each line is the ratio of the constant to the y-coefficient:

  line 1 y-intercept = 20/20 = 1

  line 2 y-intercept = k/35

We want these lines to have the same y-intercept so that they are not inconsistent. This requires ...

  k/35 = 1

  k = 35

For the equations to be consistent, we must have k = 35.

__

<em>Additional comment</em>

Linear equations are generally categorized as "consistent" or "inconsistent," and "dependent" or "independent."

Equations are "inconsistent" if there are no values of the variables that satisfy all of the equations. They are "dependent" if they describe exactly the same relation between the variables. "Consistent" equations may be "dependent" (describing the same line, as here), or "independent" (describing lines with different slopes.)

7 0
2 years ago
Any 10th grader solve it <br>for 50 points​
kkurt [141]

Answer:

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., a_{1}=A and common difference=D

The nth term can be written as

a_{n}=A+(n-1)D

pth term of given arithmetic progression is a

a_{p}=A+(p-1)D=a

qth term of given arithmetic progression is b

a_{q}=A+(q-1)D=b and

rth term of given arithmetic progression is c

a_{r}=A+(r-1)D=c

We have to prove that

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

Now to prove LHS=RHS

Now take LHS

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)

=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}

=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

5 0
3 years ago
Suppose there are two circles where the radius of one circle is twice the radius of the other circle. Each circle has an arc whe
Artemon [7]

Answer:

L=2l where l,L denote arc lengths of two circles

Step-by-step explanation:

Let l,L denote arc lengths of two circles, r,R denote corresponding radii and

\alpha _1\,,\alpha _2 denote the corresponding central angles.

So,

l=r\alpha _1 and L=R\alpha _2

This implies \alpha _1=\frac{l}{r} and \alpha _2=\frac{L}{R}

As each circle has an arc where the measures of the corresponding central angles are the same, \alpha _1=\alpha _2

\frac{l}{r}=\frac{L}{R}

As radius of one circle is twice the radius of the other circle,

R=2r

\frac{l}{r}=\frac{L}{2r}\\\frac{l}{1} =\frac{L}{2}\\L=2l

7 0
2 years ago
Determine how many solutions if correct bainliest​
PolarNik [594]

Answer:

No solutions

Step-by-step explanation:

6x+2y=20

-3x-y=-4 (by rewriting)

So

6/-3 = 2/-1≠20/-4

-2=-2≠-5

So it has no solutions

5 0
3 years ago
Help answer if u know
Setler [38]

Answer:

b = 6 / 1/4

Step-by-step explanation:

C)

7 0
3 years ago
Read 2 more answers
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