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3 years ago

8

Guadalupe wants to buy new goggles that cost $31.50. She has $4.50 and plans to save $2.25 each week. How many weeks will it tak

e her to save the money?

1 answer:

gregori [183]3 years ago

5 0

$y=2.25x+4.5\\(31.5)=2.25x+4.5\\2.25x=27\\x=12$

12 weeks.

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Read 2 more answers

Can somebody help me on 23 and 24, it’s geometry

Luda [366]

Question 23:

x = 4

DE = 44

Question 24:

x = 25

SE = 28

Step-by-step explanation:

As RS is the perpendicular bisector of DE, it will divide DE in two equal parts DS and SE

<u>Question number 23:</u>

Given

DS = 3x+10

SE = 6x-2

As the two segments are equal:

$DS = SE\\3x+10 = 6x-2$

Subtracting 10 from both sides

$3x+10-10 = 6x-2-10\\3x = 6x-12$

subtracting 6x from both sides

$3x -6x = 6x-6x-12\\-3x = -12$

Dividing both sides by -3

$\frac{-3x}{-3} = \frac{-12}{-3}\\x = 4$

Now

$DS = 3x+10\\= 3(4)+10\\= 12+10\\=22$

And

$SE = 6x-2\\= 6(4)-2\\= 24 - 2\\=22\\DE = DS+SE\\= 22+22\\=44$

<u>Question No 24:</u>

Given

DS = x+3

DE = 56

We know that:

$DS = \frac{1}{2}DE\\x+3 = \frac{56}{2}\\x + 3 = 28\\x = 28-3\\x = 25$

So

DS = $25+3 = 28$

As DS is 28, SE will also be 28

Hence,

Question 23:

x = 4

DE = 44

Question 24:

x = 25

SE = 28

Keywords: Bisector, Line segment

Learn more about line segments at:

#LearnwithBrainly

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3 years ago

If e=mc3, what does Q=?

Julli [10]

I think the answer is.e=mc3

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3 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