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Readme [11.4K]
2 years ago
12

The normal yearly growth of a plant is 60 inches. The normal growth was ten inches more than twice the amount of last year. What

was the growth of the plant last year?​
Mathematics
1 answer:
Usimov [2.4K]2 years ago
4 0

The growth of the plant last year was 25 inches if the normal growth was ten inches more than twice the amount of last year.

<h3>What is linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

The normal yearly growth of a plant is 60 inches.

Let's suppose the growth of the plant last year was x

The normal growth was ten inches more than twice the amount of last year.

From the above statement:

10 + 2x = 60

2x = 50

x = 25 inches

Thus, the growth of the plant last year was 25 inches if the normal growth was ten inches more than twice the amount of last year.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

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The value of x is 8 .

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\frac{x}{x+6}=\frac{16}{28}\\\\7 x = 4 (x + 6)\\\\7 x = 4 x + 24 \\\\3 x = 24 \\\\x = 8

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First, pull out a factor of x.

x^6+3x^5+x^4-5x^3-6x^2-2x=x(x^5+3x^4+x^3-5x^2-6x-2)

Notice that when x=-1 (which you can arrive at via the rational root theorem), you have

(-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0

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\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2

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Now this is more readily factored without having to resort to the rational root theorem. You have

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x^2-2=0\implies x^2=2\implies x=\pm\sqrt2
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