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2 years ago

6

There are seven cars parked in a row exactly the same distance from each other the first car is 31 inches from the second car th

e first car is 62 inches from the third car how far is the seventh car from the third car

1 answer:

MakcuM [25]2 years ago

8 0

Answer:

the distance of the 3rd car from the 7th car is

217 inch

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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

quester [9]

Answer:1866

Step-by-step explanation:

Given

n=200

x=118

Population proportion P= $\frac{118}{200}$ =0.59

$\alpha$ =0.005

Realiability =99%

$Z_{\frac{\alpha }{2}}=2.576$

Margin of erroe is given by $\sqrt{\frac{p\left ( 1-p \right )}{N}}$

0.03= $\sqrt{\frac{0.59\left ( 1-0.59 \right )}{N}}$

85.667= $\sqrt{\frac{N}{0.6519}}[tex]N=1865.88[tex]\approx 1866 Students$

4 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

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3 years ago

Adam drafts his favorite thing tractor on the scale drawing. The drawing scale is 3 inches for each foot. If the drawing is 20 i

dimaraw [331]

I believe it's 6.7. Not too sure though.

3 0

2 years ago

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Can someone please help me with this? it’s algebra 2

riadik2000 [5.3K]

I am big pp lol xD hxyjdrh

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2 years ago

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What is the domain of the relation?<br><br> Express your answer using inequalities.

son4ous [18]

Answer:

-6<x<0

- -

Step-by-step explanation:

make them less than or equal to signs, I don't know how to do that on a computer, I hope I am correct

point one on the left is -6, point 2 on the right is 0 (in terms of the x-axis, that's what domain is )

when there is arrows in the graph, the x/y is always in the middle.

Good luck! Hope I'm not late!

8 0

3 years ago

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