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3 years ago

4 ( 3x - 8 ) = 4 how do i solve this problem

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

3 0

4(3x-8)=4
12x-32=4
12x=36
x=3

Zepler [3.9K]3 years ago

3 0

i'd say x=3 because,

if you do 3 times 3, it equals 9. Then you subtract 8 from it, which gives you 1. And the numbers you have left over are the four and the 1. The operation you have to do for that is multiplication, and 4x1=4.

Hope this helps :)

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Clinton earns $12 per hour cutting grass. he earned a total of $162 last week cutting grass. exactly how msny hours did he spend

professor190 [17]

Answer:13.5

Step-by-step explanation:

12•13.5=162

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3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

3 years ago

Find the two square roots of 144.

spin [16.1K]

Answer:

The square root function

sqrt(144) or √144 produces a single positive value, 12.

However if you have an equation

x² = 144, then you have two possible values for x, 12 and -12.

Two ways to look at it.

x² - 144 = 0

Difference of two squares

(x+12)(x-12) = 0

x = -12, 12

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4 0

3 years ago

Read 2 more answers

X-2y=5 2x-4y=10

Nitella [24]

Step-by-step explanation:

X-2y=5 ; 2x-4y=10

• x - 2y = 5 • 2x -4y = 10

-2y = -x+5 -4y= -2x +10

y1 = ½x - 5/2 y2 = ½x -5/2

• => y1 = y2

½x -5/2 = ½x -5/2

the systems have no solution

• graph as attached

8 0

3 years ago

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Sladkaya [172]

B.
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C.
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D.
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please mark as brainliest.

3 0

3 years ago