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Katyanochek1 [597]
3 years ago
8

Square of a binomial x^2+2x+1

Mathematics
1 answer:
garik1379 [7]3 years ago
3 0
To put an equation into (x+c)^2, we need to see if the trinomial is a perfect square. 
General form of a trinomial: ax^2+bx+c
If c is a perfect square, for example (1)^2=1, 2^2=4, that's a good indicator that it's a perfect square trinomial. 
Here, it is, because 1 is a perfect square.
To ensure that it's a perfect square trinomial, let's look at b, which in this case is 2. 
It has to be double what c is.
2 is the double of 1, therefore this is a perfect square trinomial. 
Knowing this, we can easily put it into the form (x+c)^2.
And the answer is: (x+1)^2.
To do it the long way:
x^2+2x+1
Find 2 numbers that add to 2 and multiply to 1. 
They are both 1.
x^2+x+x+1
x(x+1)+1(x+1)
Gather like terms
(x+1)(x+1)
or (x+1)^2.
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Answer:

B. 5

Step-by-step explanation:

Step 1: Rewrite the equation a bit

\frac{7^\frac{3}{4}}{7^\frac{x}{8}}=\sqrt[8]{7}\\7^\frac{3}{4}^-^\frac{x}{8}=7^\frac{1}{8}

Step 2: Place a logarithm base 7 on both sides

7^\frac{3}{4}^-^\frac{x}{8}=7^\frac{1}{8}\\log_77^\frac{3}{4}^-^\frac{x}{8}=log_77^\frac{1}{8}\\(\frac{3}{4}-\frac{x}{8})log_77=\frac{1}{8}log_77\\(\frac{3}{4}-\frac{x}{8})1=\frac{1}{8}1\\\frac{3}{4}-\frac{x}{8}=\frac{1}{8}

Step 3: Solve for x

\frac{3}{4}-\frac{x}{8}=\frac{1}{8}\\\frac{6}{8}-\frac{x}{8}=\frac{1}{8}\\-\frac{x}{8}=-\frac{5}{8}\\\frac{x}{8}=\frac{5}{8}\\x=5

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Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Check al
Alla [95]

Let's rewrite each equation in the Slope-Intercept Form of the Equation of a Line. First, let's start with the main equation:


\bullet \ 5x-2y=-6 \therefore y=\frac{5}{2}x+3


Then, our options are the following:

A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \therefore y=-\frac{2}{5}x-2 \\ \\ C) \ 2x-5y=-10 \therefore y=\frac{2}{5}x+2 \\ \\ D) \ y+4=-\frac{2}{5}(x-5) \therefore y=-\frac{2}{5}x-2 \\ \\ E) \ y-4=\frac{5}{2}(x + 5) \therefore y=\frac{5}{2}x+\frac{33}{2}


For two perpendicular lines it is true that the product of its slopes is:

m_{1}m_{2}=-1


m_{1}m_{2}=-1 \\ \\ m_{1} \ is \ the \ slope \ of \ y=\frac{5}{2}x+3, \ that \ is, \ m_{1}=\frac{5}{2} \\ \\ Then, \ the \ slope \ of \ a \ perpendicular \ line \ is: \\ \\ m_{2}=-\frac{2}{5}


According to this, only A) B) and D) might be the perpendicular lines we are looking for. Notice that these lines are the same. The other condition is that the line must pass through the point (5, -4). By substituting this point in the equation, we have:

y = -\frac{2}{5}x-2 \\ \\ -4=-\frac{2}{5}(5)-2 \\ \\ -4=-2-2 \\ \\ \boxed{-4=-4} \ True!


Finally, the right answer are:

A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \\ \\ D) \ y+4=-\frac{2}{5}(x-5)

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