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3 years ago

What is the solution to 3(2x-1)=(6x+5)

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

3 0

We are given this equation and I'm assuming you want us to solve for x

$3(2x-1)=6x+5$

Use the distributive property on the left side of the equation

$6x-3=6x+5$

Subtract both sides by 6x

$-3=5$

After trying to solve for x, we have this weird equation. Since -3 does NOT equal 5 and there is no other way for us to try and solve this, the equation has no solution.

Let me know if you need any clarifications, thanks!

~ Padoru

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which of the following represents a relation that is not a function? a. x -8 -6 7 10 y 33 31 39 33 b. x -8 -6 0 3 y 33 31 39 33

boyakko [2]

X => -8 -6 -8 3
y => 33 31 39 33
is not a function because the x value -8 results to two different y-values 33 and 39.

8 0

3 years ago

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

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2 years ago

Charlie wants to order lunch for his friends. He'll order 6 sandwiches and a $3 kid's meal for his little brother. Charlie has $

yawa3891 [41]

Answer:

Step-by-step explanation:

Charlie has $27

He spent $3 for kid's meal for his little brother

Remaining money=27-3=$24

Number of Sandwiches he wants to buy = 6

Money he needs to pay for each sandwich = 24/6 = $4

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2 years ago

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djyliett [7]

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2 years ago

Read 2 more answers

Suppose you have a job teaching swimming lessons and get paid $6 an hour. You also have job as cashier and get paid $8 and hour.

VikaD [51]

This is a system of equations problem.

L = Hours of Swim Lessons
C = Hours as Cashier

L + C = 15 This is an equation for working 15 hours total.
so
L = 15-C

6L + 8C = 100 This is an equation for making at least $100.
so
6 (15-C) + 8C = 100 *You substituted 15-C for L here
90 - 6C + 8C = 100 Distribute
90 + 2C = 100
2C = 10
C = 5 You can work 5 hours as a cashier

L + C = 15
L + 5 = 15
L = 10 You can work 10 hours teaching swimming lessons.

If you want to make EXACTLY $100 you would work 5 hours as cashier and 10 hours teaching swimming lessons.

HOWEVER, the question says AT LEAST $100 and NOT MORE than 15 hours per week. Since you make more money as a cashier, any work over 5 hours will help you make over $100.

As long as you work at least 5 hours as a cashier and any remaining hours teaching swimming lessons, you will make over $100.

5 cashier, 10 swimming
6 cashier, 9 swimming
7 cashier, 8 swimming...

Mr. Pacey
JH/HS Social Studies Teacher
(but I've also helped with Math Team for JH & HS)

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2 years ago