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2 years ago

How do o solve this equation? 4(3x - 10) + 7 = 15

Mathematics

2 answers:

Rama09 [41]2 years ago

5 0

X=4 because first u distruste and add the numbers then add 33 to both side

Aleksandr-060686 [28]2 years ago

4 0

Answer:

x = 4

Step-by-step explanation:

4(3x - 10) + 7 = 15

multiply 3x and 10 by 4

12x - 40 + 7 = 15

add 40 both sides

12x + 7 = 55

subtract 7 from both sides

12x = 48

divide

x = 4

hope this helps :)

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Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

Art [367]

Answer:

Error: $lnx^2=ln 3x$ not $ln\frac{3x}{0}$

Solution:x=0 and 3

Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

$lnx^2=ln(3x)-0$

$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

4 0

3 years ago

Convert 3/11 to decimal equivalent using long division

sergey [27]

So simple... 3\ll=27\100 which equals 27.0 or 0.27

7 0

3 years ago

The two lines graphed below are parallel. How many solutions are there to the

NISA [10]

The is going to be infinite answers

7 0

2 years ago

convert the mixed numbers to improper fractions. this will get you one step closer to finding the unit rate

Kaylis [27]

It means for example instead of writing 2 2/5 You would write 12/5
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2 years ago

What is the solution of x=2+ square root x-2?

alex41 [277]

So I'm going to assume that this question is asking for non extraneous solutions, or solutions that are found in the equation and are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

$x-2=\sqrt{x-2}$

Next, square both sides:

$(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2$

Next, subtract x and add 2 to both sides of the equation:

$x^2-5x+6=0$

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

$x^2-2x-3x+6=0$

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

$x(x-2)-3(x-2)=0$

Now you can rewrite the equation as $(x-3)(x-2)=0$

Now, apply the Zero Product Property and solve for x as such:

$x-3=0\\x=3\\\\x-2=0\\x=2$

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

$2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}$

Since both solutions hold true when x = 2 and x = 3, your answer is C. x = 2 or x = 3.

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3 years ago