(01.04)Solve (-3) . 2.

Mathematics

1 answer:

jonny [76]3 years ago

7 0

Answer:

If your asking for the solution for (-3)2 the answer is -6 if that is not what your asking please clarify by commenting on this answer

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Help please; I need the right answer

Ad libitum [116K]

Answer:

Step-by-step explanation:

$\cos x+\sqrt{2}=-\cos x \\ 2\cos x = -\sqrt 2 \\ \cos x = -\dfrac{\sqrt 2}{2}\\ \\ \Rightarrow x = \pm \arccos\Big(-\dfrac{\sqrt 2}{2}\Big)+2k\pi,\quad x\in \mathbb{Z}\\ \Rightarrow x =\pm\dfrac{3\pi}{4}+2k\pi\\ \\ \\k = -1 \Rightarrow x < 0\\\\ k=0 \Rightarrow x 2\pi$

3 0

3 years ago

Read 2 more answers

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ,2. 25. Point Q is the center of dilation. Line

abruzzese [7]

The distance between points S' and S is x= 2.5 units.

<h2>Given that</h2>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ 2. 25.

Point Q is the center of dilation.

Line segment ST is dilated to create line segment S prime T prime.

The length of QT is 1. 2 and the length of QS is 2.

The length of SS prime is x and the length of TT prime is 1. 5.

<h3>We have to determine</h3>

What is x, the distance between points S' and S?

<h3>According to the question</h3>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ, 2.25.

Also, SQ = 2 units, TQ = 1.2 units, TT'=1.5, SS' = x.

Since the line ST is dilated to S'T' with the center of dilation Q, the triangles STQ and S'T'Q must be similar.

We know that the corresponding sides of two similar triangles are proportional.

So, from ΔSTQ and ΔS'T'Q.

$\dfrac{SQ}{S'Q} = \dfrac{TQ}{T'Q}\\
\\
\dfrac{SQ}{SQ+SS} = \dfrac{TQ}{TQ+TT'}\\
\\ 
\dfrac{2}{2+x} = \dfrac{1.2}{1.2+1.5}\\\rm 
\\
\dfrac{2}{2+x} = \dfrac{1.2}{2.7}\\\\ \dfrac{2}{2+x} = \dfrac{12}{27}\\\\2(27) = (2+x) 12\\\\ 54 = 24 + 12x\\
\\
12x = 54-24\\
\\
12x=30\\
\\
x = \dfrac{30}{12}\\
\\
x = 2.5$