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mote1985 [20]
3 years ago
6

What is four blank four blank four equals 4

Mathematics
1 answer:
attashe74 [19]3 years ago
3 0
4 - 4 + 4 = 4.
four minus for plus four equals four.
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You are riding the swings at a fair. What are the coordinates of your location after a rotation of 180°?
nekit [7.7K]

Answer:

(6, -3)

Step-by-step explanation:

The actual coordinates are x = -6 and y = 3

If we rotate 180 degrees and the center of rotation is at (0,0), all we need to do is invert the signal of each axis, that is, we invert the sign of the original x-coordinate and invert the signal of the original y-coordinate.

So the final x-coordinate is - (-6) = 6

And the final y-coordinate is - (3) = -3

So the coordinates will be (6, -3).

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3 years ago
57,600 at 4% compounded semiannually for 3 years
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3 years ago
Square of a binomial x^2+2x+1
garik1379 [7]
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.
3 0
3 years ago
What is the value of B Image below
Gnesinka [82]
It is 61 degrees bec u j add them then subrteact that by 189
4 0
3 years ago
Read 2 more answers
Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2
Agata [3.3K]
\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies 
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------

\bf \begin{cases}
v1=\ \textless \ -6,4\ \textgreater \ \\
v2=\ \textless \ -3,6\ \textgreater \ \\
------------\\
v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\
\qquad \qquad 42\\
||v1||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left(  \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o
8 0
3 years ago
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