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Tems11 [23]
3 years ago
9

Write a compound inequality that represents the situation. All real numbers at least -6 and at most 3.

Mathematics
2 answers:
Nikolay [14]3 years ago
4 0
All real numbers at least - 6 and at most 3:
- 6 ≤ x ≤ 3
Thank you.
nydimaria [60]3 years ago
3 0

Answer: Compound inequality will be

-6\leq x\leq 3

Step-by-step explanation:

Since we have given that

All real numbers at least -6

Let the number be x

So, According to question, it will be

x\geq -6

Now, all real numbers at most 3

So, it will be

x\leq 3

so, Compound inequality will be

-6\leq x\leq 3

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The area of a rectangle in 24 units. A diagonal drawn from one corner to another divides the rectangle into two right triangles.
nlexa [21]

Answer:

12 sq u

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, assuming the shaded triangle is one of those two triangles, its area would be 12 sq u.

Why? Because its length would be the same as the retances, as would its height, and to find the area of a triangle, you'd do L*H/2 which would be 24/2=12 sq u

(っ◔◡◔)っ ♥ Hope this helped, have a great day! ♥

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7 0
3 years ago
Which of these r-values represents the weakest correlation? A.-0.55 B -0.45 C -0.25 D -0.35
irakobra [83]
The smaller the magnitude of the correlation, the weaker the correlation.  The magnitude of -0.25 is 0.25, the smallest correlation magnitude.  
6 0
4 years ago
What is the complex conjugate of -2 + 3i?​
marta [7]
It’s 4.5 to go down to my -790
8 0
3 years ago
Could somebody help me with
JulijaS [17]

Answer:

t + 0.8 =1.5

Step-by-step explanation:

Our objective is to get to 1.5. we know that part of the distance is 0.8 but we don't know the other half (t).

7 0
3 years ago
I need help plz. Its DeMoivres Theorem
vagabundo [1.1K]

\bf \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ z=\stackrel{a}{-1}\stackrel{b}{-\sqrt{3}~i}~~ \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right)\\[-0.5em] \hrulefill\\ r=\sqrt{(-1)^2+(\sqrt{3})^2}\\ \qquad \sqrt{4}\\ \qquad 2\\ \theta =tan^{-1}\left( \frac{-\sqrt{3}}{-1} \right)\\\\ \qquad \frac{4\pi }{3} \end{cases}

\bf z=2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right]\implies z^6=\left[ 2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right] \right]^6 \\\\\\ z^6=2^6\left[ cos\left( 6\cdot \frac{4\pi }{3} \right) -i~sin\left( 6\cdot \frac{4\pi }{3} \right)\right]\implies z^6=64[cos(8\pi )-i~sin(8\pi )] \\\\\\ z^6=64[(-1)-i~(0)]\implies z^6=\stackrel{\stackrel{a}{\downarrow }}{-64}~~\stackrel{\stackrel{b}{\downarrow }}{-0i}

4 0
4 years ago
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